ROOKS'  MENTAL  ARITHMETIC, 


PHILADELPHIA: 
SOWER,   BARNES  &   CO. 


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METHODS 


TEACHING  MENTAL  ARITHMETIC, 


KEY 


THE  NORMAL  MENTAL  ARITHMETIC. 

CONTAINING   ALSO    MANY    SUGGESTIONS     AND    METHODS     FOR   ARITH- 
METICAL   CONTRACTIONS,    AND    A    COLLECTION   OF   PROBLEMS 
OF    AN    INTERESTING    AND     AMUSING    CHARACTER, 
FOR    CLASS    EXERCISE. 

BY 

EDWARD  BROOKS,  A.  M. 

PROFESSOR  OF   MATHEMATICS  IN  THE  PENNSYLVANIA  STATE  NORMAL  SCHOOL. 


SOWER,    BARNES   &   CO. 


Entered,  according  to  Act  of  Congress,  in  the  year  1860,  by 
EDWARD  BROOKS, 

In  uie  Clerk's  Office  of  the  District  Court  of  the  United  States,  in  and  for  the 
Eastern  District  of  Pennsylvania. 

HEARS  *  DUSENBERY,  STEREOTYPED, 


PEEFACE. 


Tins  little  volume  consists  of  four  distinct  parts  : 

First. — An  exposition  of  some  of  the  principles  and 
methods  of  successful  instruction  in  the  science  of  Mental 
Arithmetic. 

Second. — A  fuller  development  of  the  principles  pre- 
sented in  "The  Normal  Mental  Arithmetic/'  by  many 
remarks  and  suggestions,  and  by  the  solution  of  some  of 
the  more  difficult  problems. 

Third. — The  presentation  of  quite  a  number  of 
methods  of  numerical  computation  by  contractions,  &c. ; 
the  object  of  which  is  to  make  pupils  ready  and  accurate 
in  the  mechanical  operations  of  Arithmetic. 

Fourth. — A  collection  of  a  large  number  of  problems 
of  an  amusing  and  interesting  character,  under  the  head 
of  "Social  Arithmetic/'  to  be  used  to  awaken  interest  in 
a  class,  or  entertain  a  social  circle. 

With  regard  to  the  second  part  of  the  work,  in  which 
it  may  be  considered  a  "  Key/'  it  will  be  noticed,  that  it 
differs  from  Keys,  generally,  in  presenting  many  remarks 

165027.  (0 


11  PREFACE. 

and  suggestions,  and  also  in  solving  only  a  few  of  the 
more  difficult  problems,  so  that  even  if  a  class  of  pupils 
should  happen  to  make  use  of  it,  in  preparing  their 
lessons,  they  will  find  great  variety  still  among  the  un- 
solved problems,  for  the  exercise  of  their  own  ingenuity. 
The  author  hopes  that  the  book  may  be  found  of  value 
to  the  private  student,  to  the  teacher  of  Mental  Arith- 
metic, and  particularly  to  his  numerous  friends,  who 
have  shown  their  appreciation  of  his  former  works ;  and 
thus,  in  various  ways,  lend  assistance  to  the  great  cause 
of  the  age, — the  cause  of  Popular  Education. 


METHODS  OF  TEACHING 

MENTAL  ARITHMETIC. 

CHAPTER  I. 

ARITHMETIC  is  the  logic  of  numbers,  and  hence  its 
truths  and  principles  should  be  derived  by  logical  pro- 
cesses. The  four  logical  processes  by  which  these  truths 
and  principles  are  obtained,  are  Analysis,  Synthesis, 
Induction,  and  Deduction. 

ANALYSIS. — Analysis  is  the  process  of  resolving  that 
which  is  complex  into  its  elements.  A  watch  is  a  com- 
plex being,  consisting  of  wheels,  springs,  pins,  hands, 
&c..  Now,  if  we  take  a  watch  and  separate  it  into  these 
parts,  we  are  said  to  analyze  it.  A  house  is  analyzed 
when  we  separate  it  into  the  brick,  wood,  stone,  mortar, 
iron,  &c.,  of  which  it  is  built.  Analysis  then  meaos 
separating,  taking  to  pieces,  resolving  the  complex  into 
the  parts  of  which  it  is  composed. 

SYNTHESIS. — Synthesis  is  the  reverse  of  Analysis.  It 
is  the  process  by  which  we  form  a  complex  object  from 
simpler  objects.  Thus,  if,  after  taking  the  watch  apart 
by  Analysis,  we  put  the  different  parts  together  again, 
the  process  is  called  Synthesis.  The  building  of  a  house 
is  a  synthetic  process.  The  putting  together  of  words  to 
form  a  sentence,  is  also  a  process  of  Synthesis. 

1*  (3) 


4  METHODS   OP   TEACHING 

INDUCTION. — Induction  is  the  logical  process  by  which 
we  derive  general  truths  from  particular  ones.  Thus, 
when  we  notice  that  heat  expands  iron,  lead,  copper,  &c., 
we  infer  that  heat  will  expand  all  metallic  bodies,  and 
this  process  of  inferring  the  general  truth  is  called  In- 
duction. As  simple  as  this  logical  process  may  seem,  it 
is  of  vast  importance  in  nearly  every  department  of 
science  Its  utility  has  been  long  recognised  in  the 
Natural  and  Philosophical  sciences,  but  not  in  the  science 
of  Mathematics ;  it  is,  however,  of  very  great  importance 
even  here,  particularly  in  Arithmetic.  In  fact,  it  has 
long  been  used  in  Arithmetic,  although  apparently  un- 
consciously, and  also  in  a  slight  degree  in  Geometry  and 
Algebra.  Whenever  a  general  truth  or  principle  is 
derived  in  Arithmetic,  without  demonstration,  it  must  be 
so  derived  by  Induction.  We  have  employed  this  pro- 
cess in  Arithmetic  as  a  logical  method  of  procedure,  and 
one  which  we  claim  to  be  of  much  importance.  Its  appli- 
cation and  utility  will  be  particularly  seen  in  the  treat- 
ment of  Fractions. 

DEDUCTION. — Deduction  is  the  reverse  of  Induction. 
In  Induction,  as  has  been  seen,  we  pass  from  the  special 
to  the  general;  in  Deduction  we  reverse  this  process, 
and  pass  from  the  general  to  the  special.  Thus,  having 
derived  by  Induction  our  general  principle  that  heat  ex- 
pands all  metallic  bodies,  we  infer  that  heat  will  expand 
zinc  and  silver,  since  zinc  and  silver  are  metals.  Induc- 
tion is  an  ascending  process — we  go  up  from  the  parts 
to  the  whole }  Deduction  is  a  descending  process — we 
descend  from  the  whole  to  the  particulars.  Induction  is 
a  synthetic  process,  it  puts  together;  while  Deduction  is 
an  analytic  process,  it  takes  apart,  or  separates  the  few 
from  the  many  or  the  whole. 

INDUCTIVE  TEACHING. — There  is  a  method  of  teaching 
so  intimately  allied  to  the  logical  process  of  Induction, 
that  educators  have  given  it  the  name  of  the  Inductive 
Method.  It  consists  in  leading  the  pupil  along,  step  by 
step,  to  the  conclusion,  announcing  such  conclusion  only 


MENTAL   ARITHMETIC.  5 

after  it  is  clearly  seen  by  the  pupil.  Thus,  in  the  defi- 
nition of  Addition,  by  the  Inductive  method,  we  would, 
by  appropriate  questions,  lead  the  learner  to  a  clear  idea 
of  Addition,  and  then,  but  not  till  then,  we  would  give 
the  name  to  the  process,  and  also  the  definition  of  the 
name. 

DEDUCTIVE  TEACHING. — The  Deductive  method  of 
instruction  is  the  reverse  of  the  Inductive  method,  and 
is  so  called  because  it  resembles  deductive  logic.  By 
this  method  a  term,  and  the  definition  of  it,  are  both 
given  before  the  pupil  has  any  idea  of  the  thing  defined, 
and  then  he  is  led  to  an  understanding  of  it  by  appro- 
priate examples  and  illustrations. 

Without  entering  into  a  discussion  of  the  merits  of 
these  two  methods,  the  author  would  remark  that  he 
prefers  the  Inductive  method  for  beginners,  and  the 
Deductive  for  more  advanced  pupils.  In  the  oral  and 
mental  exercises  of  the  "Primary  Arithmetic"  the  Induc- 
tive method  has  been  employed,  and  also,  for  the  most  part, 
in  the  "  Normal  Mental  Arithmetic/'  In  the  more  ad- 
vanced departments  of  Mathematics,  the  Deductive 
method  is  preferred. 


CHAPTER  II. 

WE  have,  in  the  previous  chapter,  given  a  brief  outline 
of  the  general  principles  of  Arithmetical  Reasoning  and 
Instruction.  These  principles  will  be  found  applied  in 
the  two  little  works  of  the  author  —  THE  NORMAL 
PRIMARY  ARITHMETIC  and  THE  NORMAL  MENTAL 
ARITHMETIC. 

Arithmetic,  considered  from  the  stand-point  of  teaching, 
should  consist  of  Oral  Arithmetic,  Mental  Arithmetic, 
and  Written  Arithmetic.  It  will  be  noticed  that  we  use 
the  terms  Mental  and  Written,  instead  of  the  more  fre- 


6  METHODS    OF    TEACHING 

quently  used  ones,  "  Intellectual"  and  "  Practical." 
Our  reason  for  this  is,  that  the  terms  are  more  appro- 
priate, as  a  very  little  thought  will  show.  The  term 
"Practical"  as  applied  to  Arithmetic,  is  a  misnomer — 
all  Arithmetic  should  be  practical.  In  one  case  we  solve 
questions  mentally,  and  in  the  other  case  we  employ 
written  characters  to  aid  the  mind  in  the  computation ; 
hence  the  propriety  of  the  division  of  Arithmetic  into 
Mental  and  Written. 

ORAL  ARITHMETIC. — Oral  Arithmetic  consists  of  such 
instructions  in  the  science  and  art  of  numbers  as  should 
precede  the  use  of  a  text-book  by  the  student.  The 
learner  needs  such  instructions  for  several  reasons.  First, 
pupils  can  learn  Arithmetic  before  they  can  read,  and 
hence,  of  course,  before  they  can  use  a  book.  Secondly, 
even  with  pupils  who  can  read,  such  exercises  are  a  very 
valuable  preparation  to  the  study  of  the  subject  from  the 
text-book. 

The  chief  instrument  to  be  employed  in  these  oral 
exercises,  is  the  "Arithmometer,"  or  Numeral  Frame, 
although  books,  pens,  pencils,  grains  of  corn,  &c.,  form 
a  valuable  introduction  to  it. 

In  the  "  Normal  Primary  Arithmetic"  we  have  given 
many  suggestions  for  exercises  of  this  kind,  which  we  trust 
will  be  found  valuable.  The  teacher  can  vary  and  modify 
them  to  suit  the  capacity  and  advancement  of  the  class. 

FIRST  ARITHMETIC. — The  author  believes  that  thp 
Primary  Arithmetic  should  be  based  upon  the  following 
principles,  some  of  which  have  not  previously  been 
recognised  in  the  preparation  of  such  books. 

1st.  It  should  consist  of  both  Mental  and  Written 
exercises,  and  not  of  either  alone. 

2d.  Addition  and  Subtraction  should  be  so  presented 
in  the  Mental  Exercises  that  they  may  be  taught  simul- 
taneously, or  that  an  additive  process  should  be  immedi- 
ately reversed,  thus  giving  rise  to  a  subtractive  one. 

3d.  In  the  Mental  exercises  Multiplication  and  Divi- 
sion should  be  taught  simultaneously,  Division  being 


MENTAL    ARITHMETIC.  / 

raade  to  depend  upon  Multiplication,  as  it  logically  does, 
by  reversing  the  multiplicative  process. 

4*h.  The  pupils  should  derive  their  own  "  Multipli- 
cation Table/'  so  that  they  may  understand  its  meaning 
and  use,  being  required  to  commit  it  to  memory  after 
they  have  obtained  it. 

5th.  The  pictures  of  objects,  marks,  stars,  &c.,  should 
be  omitted,  since  the  objects  themselves  are  better  than 
their  pictures  or  the  marks,  &c.,  and  also,  since  the  pupils 
should  not  leave  the  Oral  Exercises  until  they  can  operate 
without  the  aid  of  objects  or  pictures. 

6th.  Principles  and  definitions  should  be  taught  In- 
ductively, and  the  methods  in  fractions  should  be  derived 
by  inductive  inferences  from  analytic  processes. 

MENTAL  ARITHMETIC. — Although  the  first  book,  The 
Primary  Arithmetic,  should  consist  of  a  combination  of 
Mental  and  Written  exercises,  yet  it  has  been  found 
most  advantageous  in  teaching  to  separate  them  after  the 
tirst  book,  having  a  complete  course  of  Mental  Arithmetic 
in  one  book,  and  a  complete  course  of  Written  Arithmetic 
in  another  book. 

The  Mental  Arithmetic,  it  is  believed,  should  be  based 
upon  the  following  principles  : 

1st.  It  should  be  Analytical  and  Synthetical.  Num- 
bers are  formed  by  Synthesis;  hence,  in  the  solution  of 
problems,  this  synthetic  process  must  often  be  reversed 
by  the  process  of  Analysis.  Almost  every  solution,  if 
properly  given,  involves  the  two  processes  of  Analysis 
and  Synthesis.  Since  the  synthetic  process  is  so  easy 
after  the  analysis  has  been  made,  the  two  processes  have 
been  denoted  by  the  term  Analysis. 

These  two  processes  are  included  in  the  more  general 
process  of  Comparison.  Comparison  is  more  properly 
the  reasoning  process,  Analysis  and  Synthesis  the  me- 
chanical processes. 

The  simplest  process  of  Comparison  which  involves 
Analysis,  is  where  we  compare  a  number  with  one,  a  col- 
lection with  the  unit,  since  the  relation  is  intuitivelj' 


8  METHODS   OF    TEACHING 

apprehended;  and  the  simplest  process  of  Comparison 
which  involves  Synthesis,  is  when  we  compare  the  unit 
with  the  collection.  After  the  pupil  becomes  familiar 
with  these  elementary  processes  of  Comparison,  he  begins 
to  perceive  relations  existing  between  different  collections 
of  units — numbers — and  he  should  be  taught  to  apply 
those  relations  in  the  solutions  of  problems.  This  may 
be  seen  illustrated  on  pages  42  and  43  of  the  "  Normal 
Mental  Arithmetic/' 

2d.  Fractions  should  be  treated  by  Analysis  and  Syn- 
thesis and  Induction,  or  more  briefly  by  Comparison  and 
Induction.  In  the  treatment  of  fractions  it  is  necessary 
to  obtain  methods  for  operating  upon  them,  such  as  re- 
ducing to  lowest  terms,  to  common  denominator,  &c. 
These  methods  may  be  derived  by  Induction  from  the 
results  obtained  by  the  analytic  process.  This  feature, 
which  has  not  previously  been  introduced  into  Arithmetic, 
we  deem  a  very  valuable  one,  viewed  either  from  the 
stand-point  of  science  or  teaching. 

3d.  The  problems  should  be  so  varied  that  pupils  will 
be  forced  to  think  for  themselves.  If  all  the  problems 
under  any  class  of  problems,  are  like  the  one  which  is 
solved  in  the  text-book,  the  pupils  can  take  the  solution 
given  and  apply  it  directly  to  them  all,  and  thus  no  more 
thought  is  required  than  by  the  old  method  of  working 
by  Rule.  To  prevent  this  mechanical  way  of  operating, 
we  have  taken  much  pains  to  give  great  variety  to  the 
problems. 

The  development  of  these  features,  and  also  of  others 
which  we  deem  of  much  value,  will  be  seen  by  an  exami- 
nation of  the  Key,  and  also  of  the  work  itself. 

OBJECT  OF  THE  KEY. 

This  Key  is  designed  to  show  the  manner  in  which 
Mental  Arithmetic  may  be  successfully  taught.  In  pur- 
suing this  object,  many  remarks  and  suggestions  have 
been  given,  quite  a  number  of  solutions  presented  aa 


MENTAL   ARITHMETIC.  9 

models,  and  remarks  given  for  the  solution  of  some  of 
the  more  difficult  problems.  The  object  has  not  been, 
as  in  Keys  generally,  to  solve  the  difficult  problems 
merely,  and  give  results,  but  to  suggest  to  teachers 
Methods  of  Teaching,  which  will  be  of  use  to  them  in 
the  school-room. 

When  a  solution  is  given  as  a  model  for  the  recitation- 
room,  we  head  it  with  "  SOL.  ;"  but  when  suggestions 
merely  are  given,  we  use  the  heading  "  Suo."  Remarks 
are  indicated  by  the  abbreviation  "  REM." 


CHAPTER  III. 
METHODS  OP  RECITATION. 

THE  attention  of  teachers  is  respectfully  solicited  to 
the  following  Methods  of  Recitation.  Some  of  them  are 
preferable  to  others,  but  all  may  occasionally  be  used 
with  advantage. 

COMMON  METHOD. — By  this  method  the  problems  are 
read  by  the  teacher  and  assigned  promiscuously,  the 
pupils  not  being  permitted  to  use  the  book  during  reci- 
tation, nor  retain  the  conditions  of  the  problems  by 
means  of  pencil  and  paper,  as  is  sometimes  done.  The 
pupil  selected  by  the  teacher  arises,  repeats  the  problem, 
and  gives  the  solution,  at  the  close  of  which  the  mistakes 
that  may  have  been  made  should  be  corrected  by  the 
class  or  teacher. 

SILENT  METHOD. — By  this  method  the  teacher  reads 
a  problem  to  the  class,  and  then  the  pupils  silently  solve 
it,  indicating  the  completion  of  the  solution  by  the  up- 
raised hand.  After  the  whole  class,  or  nearly  the  whole 
class,  have  finished  the  solution,  the  teacher  calls  upon 
some  member,  who  arises,  repeats  the  problem,  and  gives 
the  solution  as  in  the  former  method. 


10  METHODS   OF   TEACHING 

By  this  method  the  whole  class  must  be  exercised 
upon  every  proolem,  thus  securing  more  discipline  than 
by  the  preceding  method.  It,  however,  requires  more 
time  than  the  first ;  hence,  not  so  many  problems  can  be 
solved  at  a  recitation.  We  prefer  the  first  method  for 
advanced  pupils,  and  the  second,  at  least  a  portion  of  the 
time,  with  younger  pupils. 

CHANCE  ASSIGNMENT. — This  method  differs  from  the 
first  only  in  the  assignment  of  the  problems.  The 
teacher  marks  the  number  of  the  lesson  and  the  number 
of  the  problem,  upon  small  pieces  of  paper,  which  the 
pupils  may  take  out  of  a  box  passed  around  by  the 
teacher  or  some  member  of  the  class.  The  teacher  then, 
after  reading  a  problem,  instead  of  calling  upon  a  pupil, 
merely  gives  the  number  of  the  problem,  the  person 
having  the  number  arising,  repeating,  and  solving  it. 
By  this  method  the  teacher  is  relieved  of  all  responsibility 
with  reference  to  the  hard  and  easy  problems,  and  it  is 
also  believed  that  better  attention  is  secured  with  it.  It 
is  particularly  adapted  to  reviews  and  public  exami- 
nations. 

DOUBLE  ASSIGNMENT. — By  this  method  the  pupil 
who  receives  the  problem  from  the  teacher,  arises,  repeats 
it,  and  then  assigns  it  to  some  one  else  to  solve.  It  may 
be  combined  with  either  the  first  or  second  methods. 
The  objects  of  this  method  are  variety  and  interest. 

METHOD  BY  PARTS. — By  this  method  different  parts 
of  the  same  problem  are  solved  by  different  pupils.  The 
teacher  reads  the  problem  and  assigns  it  to  a  pupil,  and 
after  he  has  given  a  portion  of  the  solution,  another  is 
called  upon,  who  takes  up  the  solution  at  the  point  where 
the  first  stops;  the  second  is  succeeded  in  like  manner  by 
a  third ;  and  so  on,  until  the  solution  is  completed.  The 
object  of  this  method  is  to  secure  the  attention  of  the 
whole  class,  which  it  does  very  effectually.  It  is  parti- 
cularly suited  to  a  large  class  consisting  of  young  pupils. 

UNNAMED  METHOD. — By  this  method  the  teacher 
reads  and  assigns  several  problems  to  different  members 


MENTAL   ARITHMETIC.  11 

of  the  class,  before  requiring  any  solutions,  after  which 
those  who  have  received  problems  are  called  upon  in  the 
order  of  assignment  for  their  solutions.  The  advantages 
of  this  method  are,  first,  the  pupil,  having  some  time  to 
think  of  the  problem,  is  enabled  to  give  the  solution  with 
more  promptness  and  accuracy,  and  secondly,  the  neces- 
sity of  retaining  the  numbers  and  their  relations  in  the 
mind  for  several  minutes,  affords  a  good  discipline  to  the 
memory. 

CHOOSING  SIDES. — This  is  a  modification  of  the  old 
spelling  class  method,  and  is  one  calculated  to  elicit  a 
very  great  degree  of  interest.  By  it  two  pupils,  appointed 
by  the  teacher,  select  the  others,  thus  forming  two  parties 
for  a  trial  of  skill,  as  in  a  game  of  cricket  or  base  ball. 
The  problems  may  be  assigned  alternately  to  the  sides, 
by  the  teacher,  by  chance,  by  the  leaders  of  the  sides,  or 
in  any  other  way  that  may  be  agreed  upon  by  the  teacher 
and  class. 

In  regard  to  these  methods,  the  first,  second,  and  third 
are  probably  the  best  for  the  usual  recitations,  but  the 
other  methods  can  very  profitably  be  employed  with 
younger  classes,  or,  in  fact,  with  any  class,  to  relieve 
monotony  and  awaken  interest.  With  advanced  pupils 
we  prefer  the  first  method,  or  the  first  combined  with  the 
third. 

ERRORS  TO  BE  AVOIDED. 

There  is  a  large  number  of  errors  to  which  pupils  in 
every  section  of  the  country  are  liable,  a  few  of  which 
\ve  will  mention.  We  classify  them  as  errors  of  Position 
and  errors  of  Expression. 

ERRORS  OP  POSITION. — Pupils  are  exceedingly  liable 
to  assume  improper  positions  and  awkward  attitudes 
during  recitation,  such  as  leaning  on  the  desk  or  against 
the  wall,  putting  the»foot  upon  a  seat,  jamming  the  hands 
in  the  pockets,  particularly  when  the  problem  is  hard, 
playing  with  a  button,  watch  chain,  &c.  All  of  these 
2 


12  METHODS   OF   TEACHING 

faults  should  be  carefully  guarded  against,  for  reasons  so 
obvious  that  they  need  not  be  mentioned.  An  erect  and 
graceful  carriage,  aside  from  its  relation  to  health,  is  of 
advantage  to  every  lady  and  gentleman. 

ERRORS  OF  EXPRESSION. — Under  this  head  we  include 
errors  of  Articulation,  Pronunciation,  Grammar,  &c. 
There  is  quite  a  large  number  of  words  which  pupils  in 
their  haste  mispronounce,  and  also  quite  a  large  number 
of  combinations,  which  by  a  careless  enunciation  make 
ridiculous  sense,  or  nonsense.  We  will  call  the  attention 
to  a  few  of  them,  suggesting  to  the  teacher  to  correct 
these  and  others  he  may  notice. 

"And"  is  often  called  "an;"  "for"  is  called  "fur;" 
"of"  is  pronounced  as  if  the  o  was  omitted;  words  com- 
mencing with  wh,  as  when,  which,  where,  &c.,  are  pro- 
nounced as  if  spelled  "wen"  "wich"  "were"  &c. 

"Gave  him"  is  called  "aavim^"  "did  lie"  is  called 
"diddy;"  "had  he"  is  called  "haddy;"  "give  him"  is 
called  "givim;"  "give  her"  is  called  "giver;"  "which 
is"  is  often  changed  into  "witches;"  and  "how  many" 
is  frequently  transformed  into  "hominy."  "  How  many 
did  each  earn"  is  often  rendered  " hominy  did  e  churn" 

A  very  common  error,  and  one  exceedingly  difficult  to 
correct,  is  involved  in  the  following  solution  :  "  If  2 
apples  cost  6  cents,  one  apple  will  cost  the.  J  of  6  cents, 
which  are  3  cents."  Here  "the"  is  superfluous,  and 
"are"  is  ungrammatical. 

The  following  is  a  frequent  error  :  "If  one  apple  cost 
3  cents,  for  12  cents  you  can  buy  as  many  apples  as  3  is 
contained  in  12,  which  are  4  times"  The  objections  are, 
first,  3  is  not  contained  any  apples  in  12 ;  secondly,  the 
result  obtained  is  times,  when  it  should  be  apples,  or  a 
number  which  applies  to  both  times  and  apples.  The 
solution  should  be,  "You  can  buy  as  many  apples  for  12 
cents  as  3  is  contained  times  in  12,  which  are  4." 

With  regard  to  is  and  are,  it  is  not  easy  to  determine 
which  should  be  used  in  some  cases  in  Arithmetic.  I 
am  rather  inclined  to  think  that  it  would  be  better  to  use 


MENTAL   ARITHMETIC.  13 

the  singular  form  always  when  the  subject  is  either  an 
abstract  or  a  concrete  number ;  thus,  8  is  2  times  4,  or  8 
apples  is  2  times  4  apples.  But,  since  custom  sanctions 
the  use  of  "are"  with  a  concrete  number  as  a  subject,  we 
have  adhered  to  that  form.  There  is  some  authority  for 
using  "is"  in  the  "  Multiplication  Table/'  and  I  think  it 
would  be  better  if  the  singular  form  was  universally 
adopted.  We  have  adopted  the  following  rule  in  deter- 
mining the  form  of  the  verb  when  it  has  a  numerical 
subject:  If  the  idea  is  plural  use  "are;"  if  the  idea  is 
that  of  a  whole — singular — use  "is." 

Pupils  have  some  difficulty  in  knowing  how  to  read 
such  expressions  as  §|.  They  object  to  saying  "|  dol- 
lars," since  there  are  not  enough  to  make  dollars,  and 
they  also  object  to  saying  "|  of  a  dollar,"  since  there 
are  only  3  thirds  in  a  dollar.  The  correct  reading  is 
undoubtedly  the  second,  remembering  that  |  is  an  im- 
proper fraction. 

The  following  error  is  almost  universal :  2f  apples  is 
read  "2  and  3  fourth  apples"  instead  of  "2  and  3 
fourths  apples"  The  expression  " J  times"  is  sanctioned 
by  custom,  although  it  is  not  in  accordance  with  gram- 
matical principles.  It  is  rather  more  convenient  than 
the  expression  f  of  a  time,  although  evidently  a  violation 
of  the  rules  of  language. 

But  it  is  unnecessary  for  us  to  swell  this  list  larger. 
A  little  care  on  the  part  of  the  teacher  will  detect  a  large 
number  of  errors,  similar  to  those  we  have  noticed,  and 
we  suggest  that  the  attention  given  in  this  direction  will 
be  time  profitably  employed. 


KEY. 


SECTION  I. 

LESSON  I. 

A  FEW  of  the  more  difficult  problems  in  this  section 
may  be  omitted  the  first  time  in  going  over  by  a  class  of 
young  pupils.  They  are  introduced  for  the  purpose  of 
interesting  older  pupils,  that  they  may  be  induced  to 
study  "  Simple  Addition,"  in  a  Mental  Arithmetic. 

REMARK. — This  lesson  is  so  easy  that  it  is  not  neces- 
sary to  give  the  solutions  of  any  of  the  problems.  The 
15t*h  problem  is  designed  as  a  puzzle,  as  will  readily  be 
seen. 


LESSON  II. 

41.  SOL. — Since,  when  he  sold  10  none  remained, 
before  selling  these  10  he  had  10,  but  6  of  these  he 
bought,  hence  before  buying  them,  he  had  10  minus  6, 
or  4. 

54.  SOL. — If  Hiram  gave  Oliver  10  cents,  and  Oliver 

gave  Hiram  6  cents,  Oliver  had  10  —  6,  or  4  cents  more 

than  at  first,  and  Hiram  had  4  less  than  at  first ;  hence 

Oliver  had  26  +  4  or  30  and  Hiram  had  26  —  4  or  22. 

2*  (15) 


16  KEY.  [SECTION  I. 

57.  SOL.— If  A  sold  30  and  then  bought  12,  he  had 
30  — 12,  or  18  less  than  at  first,  and  if  by  selling  30 
more,  none  remained,  he  had  at  first  18  -f-  30,  or  48. 

SUGGESTION. — The  57th  may  also  be  solved  as  follows  : 

SOL.— In  all  he  sold  30  +  30,  or  60,  but  12  of  these 
he  bought,  hence  he  had  at  first  60  — 12,  or  48. 

58.  SUG.— After  A  gave  B  10  and  B  gave  A  6,  A  had 
4  less  and  B  4  more  than  at  first ;  hence  A  had  26  and  B 
34.     Then  after  B  lost  a  certain  number,  26  — 12,  or 
14,  equals  B's  number;  hence  B  lost  34  —  14,  or  20. 


LESSON   III, 

38.  SUG. — Six  times  what  they  both  have  is  90; 
hence  C  has  90 —  10,  or  80. 

40.  SUG.— After  A  gave  B  10  and  B  gave  A  20,  A 
had  10  more  and  B  had  10  less  than  at  first;  hence  A 
had  50  and  B  had  30.  Then  after  B  lost  a  certain 
number  he  had  twice  10,  or  20 ;  hence  he  lost  30  —  20, 
or  10. 

- 


LESSON  IV. 

NOTE. — The  problems  after  the  36th  may  be  assigned 
to  the  class  and  recited  as  the  other  problems,  or  to  give 
variety  to  the  exercise  the  pupils  need  not  repeat  the 
problem,  but  name  the  result  as  soon  as  the  problem  is 
announced  by  the  teacher.  It  will  be.well  for  the  teacher 
to  give  quite  a  number  of  such  problems,  as  the  exercise 
will  be  found  very  valuable. 


LESSON  V.]  KEY  17 


LESSON  V. 

81.  SOL. — Twice  a  number,  plus  8  times  the  number 
equals  5  times  the  number,  minus  4  times  the  number 
equals  once  the  number,  plus  twice  the  number  equals  3 
times  the  number. 

37  &  38.  REM. — These  problems  and  others  like  them 
may  be  made  very  interesting  to  the  pupils  by  the  teacher's 
stating  to  the  class  that  he  can  tell  the  result  which  they 
obtain,  no  matter  what  number  they  begin  with.  This 
can  always  be  done,  and  the  reason  is  as  follows :  We 
require  the  pupils  to  multiply  and  divide  until  we  reduce 
to  the  number  they  begin  with,  we  then  (not  knowing, 
of  course,  what  number  they  have)  require  them  to  add 
some  number  to  the  number  that  they  have,  and  then 
require  them  to  subtract  the  number  they  thought  of, 
which  leaves  the  number  we  told  them  to  add,  and  we 
now  know  what  number  they  have ;  we  can,  therefore, 
tell  them  to  multiply  or  divide  by  anything  we  choose, 
and  knowing  the  number  they  have,  we  can  readily  name 
the  product  or  quotient.  This  little  puzzle,  even  with 
an  advanced  class,  I  have  known  to  elicit  a  very  high 
degree  of  interest.  When  it  is  well  understood,  the 
teacher  may  show  the  class  that  they  need  not  reduce  it 
to  once  the  number  before  adding  the  number  which  he 
gives  them,  provided  they  subtract  as  many  times  the 
number  from  the  sum. 


SECTION  II. 

NOTE. — This  section  treats  of  fractions  by  using  the 
fractional  iuord^  but  not  the  expression  consisting  of  two 
figures  with  a  line  between  them.  This  is  regarded  as  a 
valuable  preparation  to  the  more  general  treatment  of  the 
following  section. 

f     ^    OF  THE    -        A 

f  UNIVERSITY    ) 


18  KEY.  [SECTION  n 


LESSON  I. 

7.  SOL.— If  1  yard  of  cloth  cost  16  cents,  1  half  of  a 
yard  of  cloth  will  cost  1  half  of  16  cents,  which  is  8 
cents. 

25.  SOL. — If  John  had  21  cents  and  gave  2  thirds  of 
them  to  Susan,  Susan  received  2  thirds  of  21  cents;  1 
third  of  21  cents  is  7  cents,  and  2  thirds  of  21  cents  are 
2  times  7  cents,  which  are  14  cents. 


LESSONS  II.  &  III. 
No  Suggestions  or  Solutions  necessary. 

LESSON  IV. 

19.  SUG.— He  sold  2  tenths  -f  3  tenths  -j-  4  tenths, 
which  are  9  tenths  of  his  number,  hence  there  remained 
10  tenths  —  9  tenths,  which  is  1  tenth  of  his  number, 
that  is,  1  tenth  of  20  sheep,  which  is  2  sheep. 

30.  SUG.— A  gave  B  20  tenths  of  a  dollar,  then  20 
tenths  +  5  tenths,  which  are  25  tenths,  equals  1  half  of 
what  I  had,  hence  I  then  had  50  tenths  of  a  dollar,  and 
at  first  56  tenths  of  a  dollar. 

38.  SUG. — One  yard  cost  4  tenths  of  a  dime,  hence 
for  12  tenths  of  a  dime  3  yards  may  be  bought. 


LESSON  V. 

NOTE. — The  previous  lessons  in  fractions  were  needed 
to  prepare  the  student  for  the  solution  of  the  problems 


LESSON  VII.]  KEY.  19 

in  this  lesson.  Some  authors  introduce  problems  similar 
to  those  in  this  lesson  previous  to  any  exercises  in  frac- 
tions, but  such  an  arrangement  is  evidently  illogical. 

47.  SuG.—The  lemons  were  worth  24  cents,  for  which 
you  could  buy  6  oranges  at  4  cents  each ;  hence  there 
remained  9  —  6,  or  3  oranges. 


LESSON  VI. 

NOTE. — The  problems  in  this  lesson  are  so  simple  that 
no  solutions  need  be  given  in  the  Key.  The  author 
would  remark  that  problems  similar  to  the  29th  and  30th 
should  frequently  be  given  to  the  class.  They  afford  an 
excellent  exercise  for  acquiring  rapidity  and  accuracy  of 
mental  computation.  The  teacher  can  readily  extem- 
porize them,  making  them  as  complicated  as  the  advance- 
in  en  t  of  the  class  may  allow. 


LESSON  VII. 

NOTE. — The  solution  given  in  the  Mental  Arithmetic 
for  the  24th  is  the  reverse  of  that  given  for  the  first  pro- 
blem of  this  lesson,  and  since  the  24th  problem  is  the 
reverse  of  the  first,  the  solution  given  seems  most  appro- 
priate. The  following  solution  to  problems  of  this  class 
is,  however,  preferred  by  some  teachers. 

24.  SOL. — In  one  there  are  three  thirds,  hence  one 
third  of  the  number  of  thirds  equals  the  number  of 
ones ;  1  third  of  6  is  2 ;  therefore,  in  6  thirds  there  are 


20  KEY.  [SECTION  n. 


LESSON  VIII. 

NOTE. — The  solution  of  the  first  problem  is  sometimes 
given  thus  :  If  3  is  one  half  of  some  number,  two  halves 
of  that  number  is  two  times  3,  or  6.  The  solution  giveu 
in  the  Arithmetic  is  perhaps  rather  more  logical. 

30.  SOL.— If  $10  is  1  third  of  6  times  the  cost  of 
the  dress,  3  thirds  of  6  times  the  cost,  or  6  times  the 
cost  of  the  dress  is  3  times  $10,  which  are  $30 ;  if  6 
times  the  cost  of  the  dress  is  $30,  once  the  cost  of  the 
dress  is  1  sixth  of  $30,  which  is  $5. 

36.  SOL.— If  1  third  of  1  half  of  some  number  is  8,  3 
thirds  of  one  half  of  that  number  is  3  times  8,  or  24 ;  if 
1  half  of  some  number  is  24,  2  halves  of  the  number 
equals  2  times  24,  which  are  48. 


LESSONS  X.  &  XI. 

NOTE. — The  method  of  solution  in  the  previous  part 
of  this  section  has  been  to  pass  from  a  collection  of  things 
or  parts  to  the  single  thing  or  part,  and  then  pass  from 
the  single  thing  or  part  to  another  collection,  thus  passing 
from  one  collection  to  another.  The  process  involves  the 
logic  of  comparison,  and  is  very  easy,  because  the  rela- 
tion between  a  unit  and  a  collection,  or  the  relation  be- 
tween a  collection  and  the  unit,  is  so  evident  that  it  is 
readily  apprehended.  The  pupils  are  now  prepared  to 
discover  relations  existing  between  different  collections 
of  units,  that  is,  between  different  numbers,  and  they 
should  be  led  to  apply  these  relations  in  the  solutions  of 
problems.  This  is  the  object  of  Lessons  X.  and  XI., 
and  we  commend  them  to  the  careful  attention  of  the 
teacher.  After  the  pupil  is  familiar  with  the  method  of 
running  to  the  unit,  it  is  a  waste  of  time  to  require  him 
to  do  so  every  time  he  wishes  to  compare  two  numbers. 


LESSON  Xlf.]  KEY.  2i 


LESSON  XII. 

NOTE. — This  we  deem  a  valuable  lesson  in  preparing 
for  fractions,  although  not  usually  introduced  into  works 
on  Mental  Arithmetic.  The  pupils  should  be  drilled 
upon  it  until  they  are  entirely  familiar  with  it.  The 
results  will,  of  course,  since  the  numbers >are  small,  be 
determined  by  inspection.  The  pupils  should  be  required 
to  frame  definitions  for  themselves  from  the  suggestive 
statements  made  in  the  lesson. 


SECTION    III. 

LESSON  I. 

1.  SOL. — In  one  there  are  f  and  in  3  there  are  3  times 
|,  which  are  |,  and  |  plus  J-  are  3.  Therefore,  in  3i 
there  are  2. 

NOTE. — One  of  the  prominent  features  of  this  book  is 
the  derivation  of  methods  in  fractions  from  the  results  of 
analytic  processes.  Unless  such  methods  are  derived  it 
would  be  necessary  to  give  the  analysis  in  full  for  every 
problem,  which  would  become  exceedingly  tedious  as  the 
problems  became  complicated.  Our  plan  then  is,  first  to 
require  the  analytic  solution,  and  after  the  pupils  are 
familiar  with  such  solution,  to  derive  a  rule,  or  method 
we  prefer  to  call  it,  by  which  the  results  may  be  obtained 
more  briefly. 

After  the  pupils  have  analyzed  a  few  of  the  problems 
in  the  first  part  of  this  lesson,  lead  them  to  see  that  they 
can  reduce  mixed  numbers  to  fractions  by  multiplying  the 
entire  part  by  the  denominator  of  the  fractional  part, 
adding  the  numerator  of  the  fractional  part  to  the  prod* 


22  KEY.  [SECTION  III. 

net,    and   using   the   result   as    the    numerator   of    the 
fraction. 

A  similar  remark  applies  to  the  problems  from  the  18th 
to  the  26th,  inclusive.  The  method  is,  Divide  the  nume- 
rator l>y  the  denominator  ;  if  there  is  a  remainder  place 
it  over  the  denominator,  and  unite  this  fraction  to  the 
'quotient  obtained  l>y  the  division. 


LESSON  II. 

15.  SOL. — It  is  evident  that  these  fractions  may  be-' 
reduced  to  12ths.  One  equals  |f,  hence  |  equals  J  of 
j|,  which  is  y^,  and  f  equals  2  times  T42,  or  -£%,  &c. 

NOTE. — Having  derived  a  method,  we  may  now  omit 
the  analysis,  and  proceed  with  this  method  as  follows. 

36.  SOL. — To  reduce  ^  to  twentieths  we  multiply  both 
numerator  and  denominator  by  10,  and  have  i,  equal  to 
•ig }  to  reduce  ^  to  twentieths  we  multiply  both  nume- 
rator and  denominator  by  5,  and  have  |,  equal  to  |g,  &c, 

46.  SOL. — If  60  cents  was  |-  of  i  of  what  he  then  had, 
\  of  -J  of  what  he  then  had  was  1  of  60  cents,  which  is 
12  cents,  and  |  of  ^  of  what  he  then  had  was  4  times  12 
cents,  or  48  cents ;  if  48  cents  was  i  of  what  he  then 
had,  |  of  what  he  then  had  equals  2  times  48  cents, 
which  are  96  cents;  hence  before  he  found  60  cents  he 
Lad  96  cents  —  60  cents,  or  36  cents.  Therefore,  &c. 

48.  ANS. — Henry  had  45  cents,  and  his  sister  had  20 
•uts,  at  first. 


LESSON  III, 

NOTE. — We  analyze  the  problems  to  the  eleventh,  and 
then  derive  a  method,  which  we  apply  in  the  solution  of 
the  problems  which  follow. 


LESSON  IV.]  KEY.  28 

15.  SOL. — To  reduce  -fa  to  fifths  we  divide  both  nu- 
merator and  denominator  by  2,  and  we  have  T65,  equal  to 
J,  &c. 

23.  SOL. — To  reduce  |  to  its  lowest  terms  we  divide 
both  numerator  and  denominator  by  the  greatest  number 
that  will  divide  them  both  without  a  remainder,  which 
number  is  4,  and  we  have  |,  equal  to  £,  &c. 

37.  SOL. — 4|  melons  =  2^4  melons.  If  2^4  melons  are 
worth  6  lemons,  ^  of  a  melon  is  worth  ^V  of  6  lemons, 
which  is  3^,  or  1  of  a  lemon,  and  |-  of  a  melon  is  worth 
f>  times  i,  or  |  of  a  lemon,  and  7  melons  are  worth  7 
\times  f ,  which  are  355,  or  8|  lemons. 


LESSON  IV. 

NOTE. — This  lesson  treats  of  the  addition  and  subtrac- 
tion of  fractions.  We  commence  by  adding  and  sub- 
tracting fractions  having  a  common  denominator,  then  by 
problems  from  10  to  1.6  suggest  the  reduction  to  a  com- 
mon denominator,  after  which  we  reduce  to  common  de- 
nominator by  the  method  derived  in  Lesson  II.,  and  then 
add  or  subtract  as  the  problem  may  require. 

17.  SOL. — i  is  equal  to  |,  and  |  is  equal, to  |;  |  plus 
|  are  |.  Therefore  the  sum  of  ^  and  |  is  |. 

47.  SOL. — If  Thomas  had  |  of  a  dollar  and  found  \  of 
a  dollar,  he  then  had  J  of  a  dollar  plus  |  of  a  dollar; 
J  =-~h  and  §  =  7%>  -h  +  T65  *=  \\-  Therefore,  &c. 

50.  SUG. — |  of  the  sum  plus  \  of  the  sum  equals  J 
of  the  sum,  which  was  $21 ;  hence  the  sum  was  §24. 

78.  SOL. — J  of  the  number  diminished  by  |  of  the 
number  equals  |  of  the  number,  which  equals  36 ;  if  | 
of  the  number  equals  36,  &c. 

NOTE. — The  pupil  may  be  led  to  see  that  when  we 
wish  to  add  two  fractions  having  a  unit  for  the  nume- 
rator, we  may  take  the  sum  of  their  denominators  for  the 
3 


24  KEY.  [SECTION  in. 

numerator  of  the  sum,  and  the  product  of  the  deno- 
minator for  the  denominator  of  the  sum ;  also,  that  the 
difference  of  two  fractions,  whose  numerators  are  a  unit, 
is  the  difference  of  their  denominators  divided  by  their 

product.     Thus,  I  +  i  =  =  &,  and  l  —  i  = 


LESSON  V. 

4.  SOL. — If  a  boy  lost  4  marbles  and  found  10,  he  then 
had  10  —  4,  which  is  6  more  than  at  first,  which  equals 
£  —  |,  or  i  of  what  he  had  at  first,  &c. 

14.  SUG. — We  find  $24  was  what  remained,  which 
equals  |  minus  |,  which  is  |  of  what  he  had  at  first,  &c. 

24.  SUG. — If  he  gave  g  of  £  away,  there  remained  the 
difference  between  |  of  ^  and  J  of  §,  which  is  |  of  §. 

28.  SUG.— |  of  f  equals  -|.  If  60  feet  is  f  of  the 
length  of  the  shadow,  diminished  by  20  feet,  60  feet 
plus  20  feet,  -which  are  80  feet,  is  f  of  the  length  of  the 
shadow,  &c. 

NOTE. — In  solving  problems  like  the  first  in  this  lesson, 
some  say  "  let  f  equal  what  he  had  at  first."  This  is 
absurd.  4  of  what  he  has  will  equal  it  whether  you  let 
it  or  not. 


LESSON  VI. 

35.  ANS. — By  dividing  the  denominator  of  j  by  2. 

36.  ANS. — By  dividing  the  denominator  of  |  by  3. 
39.  SOL. — Dividing  the  denominator  by  4  we  have  4 

times  |  equals  -•>-,  or  2^. 

59.  SOL. — 2|  =|;  3  times  f  miles  equals  8  miles; 
if  twice  the  distance  is  8  miles,  &c. 


WESSON  VIII.]  KEY.  25 


.  —  Hereafter,  when  a  fraction  is  to  be  multiplied 
by  a  number  which  will  divide  the  denominator,  insist 
upon  pupils  dividing  the  denominator,  instead  of  multi- 
plying the  numerator.  This  will  require  the  continual 
attention  of  the  teacher,  for  pupils  adhere  to  the  method 
of  multiplying  the  numerator  with  wonderful  tenacity. 


LESSON  VII. 

42.  SUG. — After  selling  f  of  f  of  a  barrel,  there  re- 
mained i  of  f  of  a  barrel,  &c. 

NOTE. — Pupils  will  naturally  solve  this  by  obtaining  | 
of  |,  and  then  subtracting  this  from  |.  The  solution 
indicated  is  much  more  concise. 


LESSON  VIII. 

NOTE. — The  problems  in  this  lesson  as  far  as  to  the 
13th  should  be  solved  like  the  1st,  then  by  means  of  the 
13th  and  14th  we  derive  a  method,  which  we  apply  to 
the  solution  of  those  which  follow. 

The  first  may  also  be  solved  thus  :  \  is  equal  to  T3j,  and 
J  of  j33  is  JTJ  ;  hence  A  of  {  is  J2.  The  solution  given  in 
the  Arithmetic  is  preferred,  since  it  involves  the  principle 
of  obtaining  a  part  of  a  fraction,  which  the  other  solution 
does  not. 

15.  SOL. — By  multiplying  the  numerators  together  for 
the  numerator  of  the  result,  and  the  denominators  for  the 
denominator,  we  find  |  of  J  to  equal  |. 

21.  SOL. — If  a  man,  owning  |  of  a  farm,  sold  £  of  it 
to  his  neighbor,  his  neighbor  received  |  of  |,  which  is 
A  of  it. 

NOTE.— The  30th  to  the  37th  we  solve  like  the  29th. 
At  the  37th  we  derive  the  method,  and  apply  this  method 
to  the  solution  of  those  which  follow. 


26  KEY.  [SECTION  in 

38.  SOL. — By  multiplying  the  numerators  together  foi 
the  numerator,  and  the  denominators  for  the  denominator 
of  the  result,  we  find  |  of  -|  equals  ^7,  &c. 

42.  SOL. — If  Johnson,  having  |  of  a  melon,  gave  |  of 
it  to  Martin,  there  remained  |  of  |  minus  |  of  |,  whict 
is  |  of  |,  which  equals  i.     Therefore,  &c. 

43.  SOL. — If  I  had  |  of  a  bushel  of  apples  and  gave  | 
of  them  away,  there  remained  |  of  them  minus  |  of  them, 
which  is  \  of  them ;  hence  there  remained  \  of  |,  which 
is  r3g  of  a  bushel. 

50.  SuG. — If  she  shared  them  with  her  schoolmates, 
they  were  divided  equally  between  herself  and  5  school- 
mates, or  6  persons ;  hence  each  received  J  of  f  =  TV 
of  a  pound. 

54.  SUG. — It  arose  \ r  of  ^  =  i,  and  then  was  ^  +  g 
=  |  of  the  first  distance  from  the  ground ;  it  fell  \  of 
i  =  2L,  and  then  was  f  —  ^  =  f  of  whole  distance 
from  the  ground. 


LESSON  IX. 

12.  ANS. — Multiply  the  dividend  by  the  fraction  ob- 
tained by  inverting  the  terms  of  the  divisor. 

NOTE. — Assign  a  few  of  the  problems  which  have 
been  solved  by  the  Analysis,  and  let  the  pupil  apply  the 
Method. 

13.  SOL.— If  a  yard  of  cloth  cost  f  of  a  dollar,  for  $12 
you  can  buy  as  many  yards  as  |  is  contained  times  in  12, 
which  are  12  X  |  —  S0?  or  2^- 

NOTE. — The  13th  problem  may  also  be  solved  by  the 
following  analytic  process.  Let  the  pupil  understand 
both. 

13.  SOL. — If  $|  will  buy  one  yard,  $i  will  buy  \  of  a 
yard,  and  $|  will  buy  5  times  -*,  which  are  f  of  a  yard, 
und  $12  will  buy  12  times  |,  which  are  20  yards. 


LESSON  XI.]  KEY.  27 

28.  SOL. — One  is  contained  in  |  |  of  a  time;  and  since 
one  is  contained  in  |  |  of  a  time,  |  is  contained  in  |  3 
times  |,  which  are  |  of  a  time,  and  |  is  contained  in  | 
5  of  |,  which  is  §  of  a  time. 

49.  ANS. — Multiply  the  dividend  by  the  fraction  ob- 
tained by  inverting  the  terms  of  the  divisor. 

NOTE. — Assign  a  few  of  the  problems  just  analyzed, 
and  require  the  class  to  solve  them  by  the  Method. 

50.  SOL. — If  a  yard  of  muslin  cost  f  of  a  dime,  for  | 
of  a  dime  you  can  purchase  as  many  yards  as  |  is  con- 
tained times  in  |,  which  is  |  X  |  =  If  >  <>r  lij- 


LESSON  X. 

16.  SOL. — i  is  2  of  f ,  and  J,  or  one,  is  3  times  ^,  which 
are  f  of  f .  It'  one  is  !  of  f ,  ^  is  I  of  f ,  which  is  T30  of 
|,  and  |  is  4  times  T3ff  =  jg,  or  f  of  f .  Therefore,  f  is 

I  <*f- 

NOTE. — This  problem  may  also  be  solved  by  reducing 
the  fractions  to  a  common  denominator,  thus : 

16.  SOL. — §  =  if*  and  |  =  yf>  and  T§  is  ^e  same 
part  of  1$  that  12  is  of  10;  and  12  is  {§,  or  f  of  10. 
Therefore,  |  is  f  of  f . 


LESSON  XI. 

24.  SOL. — There  were  as  many  persons  as  f  is  con- 
tained times  in  8,  which  are  10,  and  since  he  shared 
them,  these  included  himself;  hence  there  were  9  com- 
panions. 

26.  SuG.—  SA  =  \5.  If  ^  peaches  are  worth  5 
apples,  -i  of  a  peach  is  worth  -^  of  5  apples,  or  ^  of  an 
apple,  and  J  of  a  peach  is  worth  3  times  i,  which  are  | 
of  an  apple,  and  10  peaches  are  worth  10  times  J  of  an 
apple,  which  are  6  apples. 


28  KE?.  [SECTION  IV. 


LESSON  XII. 

1.  SOL. — If  $40  was  |  of  what  remained,  -1  of  what 
remained  was  ^  of  §40,  which  is  $10,  and  |  of  what  re- 
mained was  4  times  10,  or  $40.  After  spending  |  of  his 
fortune,  there  remained  |  —  |,  or  |  of  his  fortune,  which 
equals  $40,  &c. 

6.  SUG. — If  he  sold  ^  of  his  sheep,  ^  of  his  sheep 
remained,  which  equals  15 ;  hence  he  owned  twice  15,  01 
30.  If  -|  of  his  cows  remained,  he  sold  |  of  them,  which 
equals  10 ;  hence  he  had  at  first  15  cows. 

12.  SUG. — He  found  6  cents,  hence  he  had  12  —  6, 
or  6  cents  less  than  he  had  at  first ;  but  he  had  only  |  as 
much  as  at  first,  hence  6  cents  equals  ^  of  what  he  had 
at  first ;  therefore  he  had  at  first  4  times  6  cents,  or  24 
cents. 

19.  SUG.— She  had  at  first  50,  and  then  had  10; 
hence  she  gave  away  50  cents  —  10  cents,  or  40  cents. 

NOTE. — Let  the  pupils  be  drilled  upon  the  demon- 
strations of  the  Propositions  which  are  given  at  the  close 
of  this  lesson.  With  young  pupils  use  special  numbers 
instead  of  (?i),  until  they  can  readily  generalize,  after 
which  they  may  give  the  general  demonstration.  For 
valuable  exercises  of  this  character,  see  the  "Normal 
Primary  Arithmetic/'  pages  71,  72,  73,  and  74. 


SECTION   IV. 

NOTE. — After  the  solution  and  remarks  given  for  the 
preceding  sections,  it  will  not  be  necessary  to  solve  any 
of  the  problems  in  this  section.  It  consists  principally 
of  the  application  of  the  previous  principles  to  denominate 
numbers.  The  teacher  will  find  that  the  remarks  upon 


LESSON  I.]  '   KEY.  29 

the  tables  may  be  made  the  means  of  eliciting  much  in- 
terest in  class. 

In  the  solution  of  such  problems  as  the  8th  under  Fed- 
eral Money,  the  two  numbers  compared  must  of  course  be 
reduced  to  the  same  denomination. 

The  5th  problem  under  Apothecaries'  Weight,  and  the 
9th  under  Avoirdupois,  are  intended  to  illustrate  the 
relation  of  the  pound  and  ounce  in  the  three  different 
weights.  A  pound  of  lead  is  absolutely  heavier  than  a 
pound  of  gold,  since  the  pound  Avoirdupois  consists  of 
7000  grains,  and  the  pound  Troy  of  but  5760.  An 
ounce  of  silver,  however,  is  heavier  than  an  ounce  of 
feathers,  since  the  Troy  ounce  is  heavier  than  the  Avoir- 
dupois ounce,  as  may  be  seen  by  a  very  simple  reduction, 
that  is,  by  ascertaining  the  number  of  grains  in  an  ounce 
of  each  weight.  Thus,  in  Avoirdupois  weight  there  are 
16  oz.  in  a  pound,  hence  1  oz.  =  ^  of  7000  grs.  =  437^ ; 
but  in  a  pound  Troy  there  are  12  oz.,  hence  1  oz.  =  y1^ 
of  5760  =  480 ;  hence  an  ounce  Troy  is  42^  grs.  heavier 
than  an  ounce  Avoirdupois. 


SECTION   V. 

LESSON  I. 

IN  the  solution  of  the  problems  of  this  lesson  let  the 
pupils  observe  the  following  suggestion.  When  we 
reduce  a  collection  to  the  unit,  pass  immediately  from  this 
unit  to  the  other  collection  of  the  same  unit.  Thus,  if 
we  commence  with  a  collection  of  horses,  and  reduce  to 
one  horse,  pass  to  the  other  collection  of  horses,  instead 
of  first  reducing  some  of  the  other  numbers  to  the  unit, 
and  then  coming  back  to  the  horses. 

Many  of  these  problems  can  be  solved  without  passing 
to  the  unit,  by  employing  the  relations  of  the  numbers, 
as  is  illustrated  in  Lessons  X.  and  XL,  Section  II. 


30  KEY.  [SECTION  v. 

9.  SOL. — If  10  oxen  can  eat  4  acres  of  grass  in  6 
days,  one  ox  can  eat  4  acres  of  grass  in  10  times  6  days, 
which  are  60  days,  and  30  oxen  can  eat  4  acres  of  grass 
in  -g^  of  60  days,  which  is  2  days;  if  30  oxen  eat  4  acres 
of  grass  in  2  days,  they  will  eat  1  acre  in  |  of  2  days, 
which  is  ^  of  a  day,  and  they  will  eat  8  acres  in  8  times 
J,  which  are  |,  or  4  days. 

9.  SOL.  2d. — If  10  oxen  eat  4  acres  of  grass  in  6  days, 
30  oxen,  which  are  3  times  10  oxen,  can  eat  4  acres  of 
grass  in  ^  of  6  days,  which  is  2  days ;  and  if  they  can 
eat  4  acres  of  grass  in  2  days,  they  will  eat  8   acres, 
which  are  2  times  4  acres,  in  2  times  2  days,  which  •  are 
4  days. 

10.  SUG. — Find  how  many  men  can  perform  the  same 
piece  in  6  days,  and  then  to  perform  a  piece  3  times  as 
large  will  require  3  times  as  many  days. 

16.  SOL. — If  3  oranges  are  worth  9  cents,  one  orange 
is  worth  ^  of  9  cents,  which  is  3  cents,  and  8  oranges,  or 
4  melons,  are  worth  8  times  3  cents,  which  are  24  cents ; 
if  4  melons  are  worth  24  cents,  one  melon  is  worth  i  of 
24  cents,  or  6  cents,  and  10  melons  are  worth  10  times 
6  cents,  which  are  60  cents. 

NOTE. — This  problem  can  also  be  solved  by  com- 
mencing with  the  melons,  thus : 

16.  SOL. — If  4  melons  are  worth  8  oranges,  one  melon 
is  worth  |  of  8  oranges,  which  is  2  oranges,  and  10 
melons  are  worth  10  times  2  oranges,  which  are  20 
oranges ;  if  3  oranges  are  worth  9  cents,  one  orange  is 
worth  -1  of  9  cents,  which  is  3  cents,  and  20  oranges  are 
worth  20  times  3  cents,  which  are  60  cents. 

23.  SOL. — If  A  can  do  as  much  work  in  2  days  as  B 
can  in  4  days,  or  C  in  6  days,  then  B  can  do  as  much  in 
4  days  as  C  can  in  6  days,  and  B  can  do  as  much  in  3 
times  4  days,  or  12  days,  as  C  can  in  18  days,  which  are 
3  times  6  days. 

25.  SOL. — If  A  can  do  3  times  as  much  in  a  day  as  B, 
and  B  can  do  twice  as  much  in  a  day  as  C,  then  A  can 


LESSON  II.]  KEY.  31 

do  3  times  twice  as  much  as  C,  which  arc  6  times  as 
much  as  C ;  hence  A  can  do  as  much  in  one  day  as  C 
does  in  6  days,  and  A  can  do  as  much  in  £  of  a  day  as  C 
can  in  one  day,  and  A  can  do  as  much  in  |,  or  f  of  a  day, 
as  C  can  in  4  days. 

27.  SUG. — Find  how  long  it  will  take  8  boys  to  do  ^ 
of  it,  and  then  how  long  it  will  take  8  —  3,  or  5  boys,  to 
do  the  other  half. 

28.  SOL.— If  9  men  build  10  rods  of  wall  in  8  days, 
they  can  build  20  rods,  which  are  2  times  10  rods,  in  2 
times  8  days,  which  are  16  days,  and  they  can  build  \  of 
it  in  i  of  16  days,  which  is  4  days,  and  |  of  it,  what 
remains,  in  3  times  4  days,  or  12  days;  if  9  men  can 
build  |  of  the  wall  in  12  days,  ^  of  the  number,  which 
remains  when  |  leave,  can  build  it  in  3  times  12  days, 
which  are  36  days,  and  the  20  rods  will  be  built  in  4  -f- 
36,  or  40  days. 

29.  SOL. — If  a  measure  of  flour  make  5  four-cent 
loaves,  it  will  make  5  times  4,  or  20  one-cent  loaves ;  and 
if  it  will  make  20  one-cent  loaves,  it  will  make  i  of  20, 
or  10  two-cent  loaves. 


LESSON  II. 

2.  SUG. — After  finding  the  number  of  pupils  to  be 
13,  we  find  the  number  of  questions  by  multiplying  2  by 
13,  and  then  adding  26,  or  by  multiplying  4  by  13. 

7.  SOL. — The  difference  between  having  10  more  and 
30  more,  is  20,  and  the  difference  between  having  2  times 
as,  many  as  Robert  and  4  times  as  many  as  Robert,  is 
twice  as  many  as  Robert ;  hence  twice  Robert's  number 
equals  20,  and  Robert's  number  equals  |  of  20,  or  10,  &c. 

7.  SOL.  2d. — By  the  first  condition,  twice  Robert's 
equals  Morris'  plus  10  ;  by  the  second  condition,  4  times 
Robert's  equals  Morris'  plus  30 :  hence  4  times,  minus  2 
times,  or  2  times  Robert's  equals  30  — 10,  or  20,  &c. 


82  KEY.  [SECTION  v. 

10.  SUG. — We  find  the  cost  of  one  orange  is  4  cents, 
and  the  cost  of  one  apple  is  2  cents,  the  difference  between 
the  cosfc  of  each  is  2  cents,  the  difference  between  the 
cost  of  all  is  18  cents ;  hence  there  were  as  many  of  each 
as  2  is  contained  times  in  18,  which  are  9. 

12.  SOL. — The  difference  between  having  8  more  and 
12  less  is  20,  and  the  difference  between  having  6  times 
iis  many  as  Henry,  and  2  times  as  many  as  Henry,  is  4 
times  as  many  as  Henry;  hence  4  times  Henry's  number 
equals  20,  &c. 

13.  SOL. — If  one  girl  received  3  apples,  and  one  boy 
received  4  apples,  one  girl  and  one  boy  received  3  apples 
plus  4  apples,  which  are  7  apples;  and  since  there  were 
the  same  number  of  each,  and  all  received  28   apples, 
there  were  as  many  of  each  as  7  is  contained  times  in  28, 
which  are  4. 

17.  SOL. — If  one  boy  receives  2  cents,  3  boys  will  re- 
ceive 3  times  2  cents,  which  are  6  cents,  and  one  girl  and  3 
boys  will  receive  4  cents  plus  6  cents,  which  are  lO  cents, 
and  they  all  received  60  cents ;  hence  there  were  as  many 
times  one  girl  and  3  boys  as  10  is  contained  times  in  60, 
which  are  6 ;  hence  there  were  6  girls,  and  6  times  3,  or 
18  boys. 

18.  SUG. — We  find  the  difference  between  the  cost 
of  2  apples  and  one  orange  is  2  cents,  and  the  difference 
between  the  cost  of  all  is  10  cents;  hence  there  were  as 
many  times  one  orange  and  2  apples  as  2  is  contained 
times  in  10,  which  are  5;  hence  there  were  5  oranges, 
and  5  times  2,  which  are  10  apples. 


LESSON  III. 

6.  SOL. — Three  times  a  certain  number  equals  |  of 
the  number,  which  increased  by  |  of  the  number  equals 
y  of  the  number,  which  by  a  condition  of  the  problem 
equals  22,  &c. 


LESSON  III.]  KEY.  33 

12.  Suo. — We  find  that  |  of  the  height,  plus  10  feet, 
equals  |  of  the  height ;  then  |  of  the  height,  minus  |  of 
the  height,  which  is  |  of  the  height,  equals  10  feet,  £c. 

16.  SOL. — If  2  times  a  number,  plus  6,  equals  3  times 
the  same  number,  plus  2,  then  3  times  the  number,  minus 

2  times  the  number,  which  is  once  the  number,  equals  6 
minus  2,  which  is  4. 

19.  SOL. — If  4  times  A's  age,  minus  10  years,  equals 

3  times  A's  age,  plus  10  years,  4  times  A's  age  equals  3 
times  A's  age,  plus  20  years ;  and  if  4  times  A's  age 
equals  3  times  A's  age,  plus  20  years,  4  times  A's  age, 
minus  3  times  A's  age,  which  is  once  A's  age,  equals  20 
years. 

•  19.  SOL.  2d. — If  4  times  A's  age,  minus  10  years, 
equals  3  times  A's  age,  plus  10  years,  4  times  minus  3 
times,  or  once  A's  age  =  10  -f-  10,  or  20  years. 

23.  SUG. — After  spending  |  of  what  he  borrowed, 
there  remained  J-  of  what  he  borrowed,  that  is,  i  of  f , 
which  is  |  of  Emily's  money,  which  is  §20,  &c. 

24.  SUG. — If  the  thief  spent  f   of  what   he  stole, 
there  remained  -|  of  what  he  stole,  that  is,  -i  of  |  of 
Harry's  money,  which  is  1  of  Harry's ;  then  J  of  Harry's 
money,  what  was  stolen,  minus  %  of  his  money,  what  was 
given  back,  equals  §  of  Harry's  money,  which  equals 
$20,  &c. 

25.  SUG. — Two  times  a  number  plus  10,  equals  3 
times  (the  number  plus  2),  which  is  3  times  the  number 
plus  6;   then  3  times  the  number,  minus  2  times  the 
number,  which  is  once  the  number,  equals  10  minus  6, 
or  4,  &c. 

26.  SUG.— If  the  thief  had  spent  f  of  what  he  stole, 
there  remained  i  of  what  he  stole,  that  is,  1  of  f  of 
Baldwin's  money,  which  is  -£  of  his  money;  then  ^  of 
his  money,  what  Baldwin  had  remaining,  minus  ^  of  his 
noney,  which  is  ^  of  his  money,  equals  §30,  &c. 


34  KEY.  [SECTION  V. 


LESSON  IV. 

7.  SOL. — If  ^  of  the  longer  piece  equals  the  shorter, 
then  |  of  the  longer  piece,  plus  ^  of  the  longer  piece, 
which  is  the  shorter  piece,  equals  |  of  the  longer  piece, 
which,  by  the  condition  of  the  problem,  equals  36  feet. 
If  |  of  the  longer  equals  36  feet,  ^  of  the  longer,  which 
is  the  shorter,  equals  \  of  36  feet,  which  is  9  feet,  &c. 

11.  SUG. — We  find  the  sum  of  the  numbers  equals 
15.  If  3  times  the  smaller  number  equals  twice  the 
greater,  once  the  smaller  equals  |  of  twice  the  greater, 
which  is  |  of  the  greater ;  then  f  of  the  greater  number, 
which  is  the  smaller,  plus  |  of  the  greater,  which  is  | 
of  the  greater  number,  equals  15,  &c. 

14.  SOL. — Twice  what  the  first  has  equals  what  the 
second  has,  and  3  times  what  the  first  has  equals  what 
the  third  has }  then  once  what  the  first  has,  plus  twice 
what  the  first  has,  which  is  what  the  second  has,  plus  3 
times  what  the  first  has,  which  is  what  the  third  has, 
equals  6  times  what  the  first  has,  which  equals  36 
apples,  &c. 

17.  SOL. — If  B  earned  twice  as  much  as  C,  then  twice 
what  C  earned  equals  what  B  earned,  and  if  A  earned 
twice  as  much  as  B,  2  times  twice  what  C  earned,  which 
is  4  times  what  C  earned,  equals  what  A  earned,  &c. 

21.  SUG. — The  difference  between  4  times  the  smaller 
and  once  the  smaller,  which  is  3  times  the  smaller,  equals 
the  difference  between  the  two  numbers,  which  is  27. 

25.  SOL. — If  there  are  |  as  many  sheep  as  hogs,  |  of 
the  number  of  hogs  equals  the  number  of  sheep ;  and  if 
chere  are  |  as  many  cows  as  hogs,  |  of  the  number  of 
hogs  equals  the  number  of  cows,  &c. 

26.  SOL. — Since  |  of  the  length  in  the  water  equals 
the  length  in  the  mud,  |  of  the  length  in  the  water, 
minus  |  of  the  length  in  the  water,  which  is  \  of  the 


LESSON  V.]  KET.  35 

length  in  the  water,  equals  10  feet;  |  of  the  length  in 
the  water,  which  is  the  length  in  the  mud  =  3  times  10 
feet,  or  30  feet,  and  |  Of  the  length  in  the  water  =  4 
X  10  =  40  feet,  which  is  f  of  the  length  in  the  air,  &c. 


LESSON  V. 

5.  SUG.— Before  Reuben  found  $9,  they  had  $15  — 
$9  —  $36 ;  hence  each  at  first  had  4  of  $36,  which  is 
SIS. 

8.  SUG. — They  both  lost  10  cents;  hence  before  losing 
any  they  had  22  cents  plus  10  cents,  which  are  32  cents  , 
and  each,  therefore,  found  i  of  32  cents,  which  is  16 
cents. 

11.  SOL. — By  a  condition  of  the  problem,  3  times 
Harry's,  plus  5  years,  equals  Harvey's  age,  which,  added 
to  Harry's  age,  equals  4  times  Harry's,  plus  5  years, 
which  is  45  years.  If  4  times  Harry's  age,  plus  5  years, 
equals  45  years,  4  times  Harry's  age  equals  45  years, 
minus  5  years,  which  is  40  years,  &c. 

15.  SOL. — Three  fifths  of  the  larger  piece,  plus  5  feet, 
equals  the  shorter,  which,  added  to  the  larger,  equals  | 
of  the  larger,  plus  5  feet,  which  equals  45  feet,  &c. 

19.  SOL. — If  he  walked  5  miles  further  the  second 
day  than  the  first,  and  10  miles  further  the  third  day 
than  the  second,  then  once  the  distance  he  walked  the 
first  day,  plus  5  miles,  equals  the  distance  he  walked  the 
second  day,  and  once  the  distance  he  walked  the  first 
flay,  plus  5  miles,  plus  10  miles,  which  is  once  the  dis- 
tance he  walked  the  first  day,  plus  15  miles,  equals  the 
distance  he  walked  the  third  day,  &c. 

21.  SUG. — We  find  that  7  times  the  cost  of  the  hat, 
plus  $8,  equals  878;  hence  7  times  the  cost  of  the  hat 
equals  $78  —  $8  =  $70,  &c. 
4 


36  KEY.  [SECTION  vi. 

23.  SUG. — A  earned  |  of  |  a.s  much  as  C,  which  is 
I  as  much  as  C ;  then  f  of  what  C  earned,  which  is  what 
A  earned,  plus  |  of  what  C  earned,  which  is  what  B 
earned,  plus  4  Of  what  C  earned,  which  is  what  C  earned, 
equals  |  of  what  C  earned,  which  is  $108.     If  |  of  what 
C  earned  equals  $108,  \  of  what  C  earned  equals  i  of 
$108,  which  is  $12,  &c. 

24.  SUG. — |  of  the  number  of  sheep,  plus  10,  which 
is  the  number  of  cows,  plus  |  of  the  number  of  sheep, 
plus  10,  which  is  the  number  of  horses,  plus  |  of  the 
number  of  sheep,  equals  y  of  the  number  of  sheep, 
plus  20,  which  equals  42 ;  hence  y  of  the  number  of 
sheep  equals  42  —  20  =  22,  &c. 


SECTION  VI. 

IT  will  be  noticed  that  the  Section  on  Per  Cent,  and 
Interest  is  not  placed  near  the  latter  part  of  the  book,  as 
is  the  custom  of  most  authors.  There  are  two  reasons 
for  this;  first,  these  subjects  are  much  easier  than  some 
which  follow  them,  and  secondly,  being  of  a  more  prac- 
tical character,  they  should  be  inserted  as  soon  as  the 
pupil  is  prepared  for  them. 

It  will  be  found,  also,  that  the  subject  of  per  cent,  is 
more  thoroughly  treated  than  in  any  previous  work  of 
this  kind.  A  large  collection  of  interesting  problems, 
involving  novel  and  useful  combinations,  has  been  intro- 
duced, by  which  original  thought  on  the  part  of  the  pupil 
will  be  secured. 

ERRORS  TO  BE  AVOIDED. — In  the  solutions  of  pro- 
blems in  this  section,  teachers  should  be  particular  that 
the  words  per  cent,  and  cents  are  not  used  synonymously . 
Learners,  somehow,  get  the  idea  that  per  cent,  has  some 


LESSON  II.]  KEY.  37 

very  intimate  connection  with  cents,  and,  unless  guarded, 
•will  use  the  one  for  the  other. 

The  expression,  "  5  per  cent,  on  a  hundred/'  is  fre- 
quently used  by  learners.  This,  of  course,  is  equivalent 
to  "  5  on  a  hundred  on  a  hundred/'  the  absurdity  of 
which  is  evident. 

Another  error  is  the  expression,  "  25  per  cent,  equals 
J."  This  is  not  exactly  true.  That  25  per  cent,  of  any- 
thing is  equal  to  |  of  that  thing,  is  true;  but  it  is, not 
true  that  "25  on  a  hundred  equals  |." 


LESSON  I. 

2.  REM. — This  is  designed  for  four  different  problems, 
the  object  of  the  author  being  to  economize  space. 

7.  SOL. — 8|  per  cent,  equals  *£  per  cent.     A  gain 
of  ~g5  per  cent,  is  a  gain  of  \5  on  100.     If  on  100  the 
gain  is  2g5,  on  1  it  is  TJ<j  of  %f,  which  is  -/fa,  or 
Therefore,  at  a  gain  of  8|  per  cent.,  -fa  of  the  cost  equa 
the  gain. 

12.  SUG.— Find  20  per  cent,  of  850,  and  then  find 
20  per  cent,  of  the  remainder,  and  subtract ;  or,  find  80 
per  cent,  of  $50,  and  then  80  per  cent,  of  this. 

16.  SUG. — Each  yard  cost  80  cents;  hence  the  gain 
on  each  yard  was  j1^,  or  y1^  of  30  cents. 


LESSON  II. 

6.  SUG. — We  find  he  lost  -|  of  the  cost,  but  the  cost 
was  100  per  cent,  of  itself;  hence  he  lost  |  of  100  per 
cent.,  which  was  33  *  per  cent. 

HEM. — It  will  be  observed  that  we  do  not  use  the  price 
of  the  horse  in  the  solution ;  there  are  several  problems 


88  KEY.  [SECTION  vi. 

of  this  character  in  the  work,  the  object  being  to  exer- 
cise the  judgment  of  the  pupil  as  to  the  number  of 
essential  conditions. 

8.  SUG. — This  problem  usually  gives  rise  to  consider- 
able discussion,  some  obtaining  20,  and  others  25  per 
cent.,  for  the  answer.    The  difference  arises  in  computing 
the  per  centage  upon  the  cost  of  10  cows,  or  upon  the 
cost  of  8  cows,  the  number  sold.     If  the  rate  per  cent, 
is  reckoned  upon  10,  the  result  is  20,  if  upon  8  it  is  25. 
It  is  evident  that  the  latter  is  correct,  since  a  merchant 
reckons  the  gain  of  a  sale,  not  upon  the  value  of  his 
whole  stock,  but  upon  the  value  of  the  goods  sold. 

9.  SOL. — $25  is  100  per  cent,  of  $25 ;  and  since  $25 
is  100  per  cent,  of  $25,  $5,  which  is  £  of  $25,  is  £  of 
100  per  cent.,  which  is  20  per  cent  of  $25. 

12.  SUG. — After  losing  20  per  cent,  there  remained 
80  per  cent;  then  selling  25  per  cent.,  there  remained 
T7c>%'  or  I  °f  ^O  per  cent.,  which  is  60  per  cent. 

13.  SUG. — Here  are  several  distinct  problems.     We 
will  solve  the  first  and  last. 

SOL.  of  1st. — J  is  100  per  cent,  of  j ;  and  since  |  is 
100  per  cent,  of  |,  |,  which  is  ^  of  ^,  is  4  of  100  per 
cent.,  which  is  50  per  cent,  of  ^. 

SOL.  of  the  last. — f  is  100  per  cent,  of  §;  hence  ^  is 
?  of  100  per  cent.,  which  is  50  per  cent,  of  f ,  and  f ,  or 
1,  is  3  times  50  per  cent.,  which  are  150  per  cent,  of  f ; 
and  i  is  i  of  150  per  cent,  of  |,  which  is  30  per  cent, 
of  f ,  and  |  is  3  times  30  per  cent.,  which  are  90  per 
cent,  of  |. 

15.  SUG.— f  of  $6  =  $4 ;  £  of  $50  equals  840.  If 
$4  is  twice  some  per  cent,  of  $40,  once  that  per  cent,  of 
$40  equals  ±  of  $4,  which  is  $2,  and  $2  is  5  per  cent,  of 
$40.  Therefore,  f  of  $6  is  twice  5  per  cent,  of  £  of  $50. 

24.  SUG. — $12  equals  f  of  the  cost,  from  which  we 
can  find  the  cost  and  the  loss  if  sold  for  $8. 


LESSON  III.]  KEY.  39 


LESSON    III. 

7.  SUG. — Find  the  cost  as  in  problem  (1),  and  then 
what  he  would  have  received,  at  a  gain  of  25  per  cent. 

12.  SUG. — Find  the  value  of  the  horse  and  carriage, 
then  the  gain  if  sold  for  $300,  and  then  the  gain  per 
cent. 

14.  SOL. — If  on  the  first  he  gained  25  per  cent.,  then 
»?050,  or  A  of  the  cost,  equals  the  gain,  which,  added  to 
|  of  the  cost,  equals  f  of  the  cost,  which  equals  $15 ;  if 
J  of  the  cost  equals  $15,  \  of  the  cost,  which  is  the  gain, 
equals  1  of  $15,  which  is  $3.  If  on  the  second  book  he 
lost  25  per  cent.,  then  T205^,  or  \  of  the  cost,  equals  the 
loss,  which,  subtracted  from  |  of  the  cost,  equals  |  of  the 
cost,  which  equals  $15;  if  |  of  the  cost  equals  $15,  -J 
of  the  cost,  which  is  the  loss,  equals  \  of  $15,  which  is 
$5.  Since  on  one  he  gained  $3,  and  on  the  other  he  lost 
§5,  he  lost  $5  minus  $3,  which  is  $2,  by  the  transaction. 

NOTE. — The  14th  problem  is  sometimes  solved  by 
finding  the  cost  of  both  books,  then  the  gain  on  the  first 
and  loss  on  the  second,  and  then  the  difference  of  these. 
This  solution  is  longer  than  the  one  given,  and  no  clearer. 

18.  REM. — This  problem  usually  gives  rise  to  consi- 
derable discussion ;  some  hold  that,  on  the  stove  bought, 
he  gained  $10,  thus  considering  $30  |  of  the  value,  and 
hence  estimating  the  gain  upon  the  value  of  the  stove; 
others  say  that  the  gain  on  the  stove  bought  is  $7£,  that 
is,  25  per  cent,  of  $30,  thus  estimating  the  gain  upon 
the  money  expended.     The  latter  view  is  undoubtedly 
the  correct  one,  since  in  a  business  transaction  gain  is 
always  estimated  upon  the  money  employed  in  the  pur- 
chase, and  not  upon  the  value  of  the  article.     In  this 
way  both  the  buyer  and  seller  may  gain. 

19.  SUG. — We  find  J-  of  what  she  received  for  the 
painting  equals  J  of  the  cost;  hence  she  gained  i  of  the 
cost,  or  20  per  cent. 

4* 


40  KEY.  [SECTION  vi 


LESSON  IV. 

1.  SUG. — $20  equals  T20^,  or  ^  of  the  cost;  hence  the 
cost  is  §80. 

5.  SOL. — If  4  is  10  per  cent,  of  some  number,  then 
4  is  -j1^,  or  JIQ  of  some  number,  and  jg  of  that  number 
is  10  times  4,  which  are  40. 

7.  SOL. — If  80  is  25  per  cent,  less  than  some  number, 
then  30  is  T2y%,  or  ^  less  than  some  number;  hence  30 
is  |  of  some  number,  &c. 

8.  SOL. — If  he  gained  820  by  selling  for  20  per  cent. 
above  its  value,  by  selling  for  10   per  cent,  above  its 
value,  which  is  J  of  20  per  cent.,  he  would  have  gained 
1  of  $20,  which  is  810. 

16.  SUG. — If  A's  money  is  y2^,  or  |  more  than  B's, 
then  |  of  B's  money  equals  A's  money,  and  the  differ- 
ence between  A's  money  and  B's  money  is  \  of  B's 
money;  and  since  f  of  B's  money  equals  A's  money, 
this  difference  is  -I  of  A's,  or  20  per  cent,  of  A's  money. 

21.  SUG. — $8  is  20  per  cent.,  or  1  of  the  cost;  hence 
the  cost  is  $40,  and  therefore  he  can  buy  10  yards. 


LESSON  V. 

These  problems  are  so  easy  that  it  is  not  necessary  to 
give  the  solution  of  many  of  them.  We  solve  the  21st 
as  a  model  for  those  which  follow. 

21.  SOL.— At  5  per  cent.  T^  of  the  principal  equals 
the  interest  for  one  year,  and  for  6  years  6  times  T|^, 
which  are  -j3^,  or  -^  of  the  principal  equals  the  interest ; 
TJa  of  $60  is  $6,  and  73^  of  $60  equals  3  times  §6,  which 
are  $18. 


ON  VI.]  KEY.  41 


LESSON  VI. 

In  finding  the  amount,  the  pupil  may  add  the  interest 
to  the  principal,  or  he  may  obtain  it  by  the  method  given 
in  the  solution  in  the  Mental  Arithmetic.  We  prefe- 
the  latter  method,  since  it  prepares  the  pupil  to  find  the 
principal  when  the  amount,  rate  per  cent.,  and  time  an* 
given. 

When  the  pupils  are  familiar  with  the  method  of  find- 
ing the  part  of  the  principal  which  equals  the  interest,  as 
given  in  Lesson  V.,  page  107,  it  may  be  abbreviated,  as 
is  indicated  in  the  following  solution. 

13.  SOL. — For  8  years,  at  5  per  cent.,  y*^,  or  f  of  the 
principal  equals  the  interest,  which,  added  to  f  of  the 
principal,  equals  \  of  the  principal,  which  equals  the 
amount;  I  of  $500  equals  8100,  and  f  of  8500  equals 
7  times  §100,  which  are  §700.  Now,  A's  share  is  6 
times  B's,  which,  added  to  B's,  equals  7  times,  which 
equals  8700,  &c. 

16.  SUG. — We  find  A's  interest  equals  840,  and  since 
A's  fortune  is  |  of  B's,  A's  interest  is  |  of  B's  interest; 
hence  840  is  |  of  B's  interest,  and  |  of  B's  interest  is  4 
times  840,  or  "§160. 

17.  SUG. — Find  C's  amount,  and  this  is  |   of  D's 
amount,  from  which  D's  amount  can  be  readily  obtained. 

18.  SUG. — Find  A's  interest,  and  this  is  f  of  B's, 
from  which  find  B's,  and  then  subtract  one  from  the 
other. 

A  shorter  method  is  the  following :  Since  A  has  |  as 
much  as  B,  A  has  -i  of  B's  more  than  B,  which  is  ^  of 
8400 ;  hence  f  of  8200  =  880  is  the  difference. 

19.  HEM. — Some  find  the  principal  of  each,  and  then 
the  interest  of  each  principal.     It  is  better,  however,  to 
find  the  interest  of  81200,  and  then  divide  this  interest 
according  to  the  conditions  of  the  problem. 


KEY  [SECTION  vi 


LESSON  VII. 

15.  SUG.— We  find  the  principal  to  be  $600,  which 
./ivided  according  to  the  conditions  of  the  problem,  give? 
$200  for  A's,  and  §400  for  B's  fortune. 

17.  SUG.— We  find  the  sum  of  I  of  A's  and  |  of  B's 
fortune  to  be  S800 ;  hence,  since  ^  of  A's  equals  -i  of 
B's,  i  of  $800,  or  $100,  equals  i  o~f  A's,  and  also  |  of 
B's  fortune,  &c. 

18.  SUG. — This  problem  is  usually  solved  by  finding 
how  much  Martin  owes,  and  then  how  much  money  is 
due  him,  and  then  taking  this  difference.     The  following 
method,  however,  is  much  shorter. 

We  find  that  1  of  the  principal  equals  the  interest, 
and  the  difference  of  the  interest  of  the  money  he  owes 
and  the  money  due  him,  is  $30 ;  hence  $30  is  i  of  what 
is  due  above  what  he  owes. 


LESSON  VIII. 

7.  REM. — Pupils  in  solving  these  problems  usually  say, 
as  in  the  previous  examples,  J  of  the  principal  equals 
the  interest.  This  is  not  precise.  It  should  be,  1  of  the 
present  worth  equals  the  discount. 

13.  SUG. — Pupils  usually  find  the  present  worth,  and 
subtract  it  from  the  debt  to  find  the  discount.  The 
shorter  way  is  to  find  the  value  of  the  part  of  the  pre- 
sent worth  which  equals  the  discount,  thus  :  ^90  of  the 
present  worth  equals  the  discount,  and  f  g  of  the  present 
worth  equals  $580,  ^  of  the  present  worth  equals  ^  of 
$580,  which  is  $20,  and  -^  of  the  present  worth,  which 
is  the  discount,  equals  9  times  20,  or  $180. 

20.  SUG. — This  means  that  3  times  B's  equals  A's; 
hence  4  times  A's  plus  A's,  which  is  5  times  A's,  equals 
$3000. 


LESSON  IX.]  KEY.  43 

22.  SUG. — The  principal  is  §1500.  If  the  house  cost 
•J  as  inuch  as  the  farm,  |  of  the  cost  of  the  farm  equals 
twice  the  cost  of  the  house;  hence  2^  times,  or  |  of  the 
cost  of  the  house  equals  $1500,  &c. 

24.  SUG. — We  find  that  ^  of  the  principal  equals  the 
interest,  and  |  of  the  principal  equals  the  amount;  hence 
$  of  first  interest  equals  8270,  and  the  first  interest 
equals  $180,  which  is  ^  of  the  principal;  hence,  the 
principal  equals  §360. 

Or,  since  ^  of  the  first  principal  equals  the  interest, 
and  |  of  this  interest  equals  the  amount,  •§•  of  -J,  or  |  of 
the  cost  of  the  horse,  &c.,  equals  §270;  hence  |  of  the 
cost  equals  §360. 


LESSON  IX. 

15.  HEM.  —  This  problem  can  also  be  solved  by  the 
following  method  :  A  principal  will  gain  twice  itself  in 
200  years,  at  one  per  cent.,  and  at  40  per  cent,  in  fa  of 
200  years,  or  5  years. 

17.  SOL.  —  At  5  per  cent,  for  one  year,  T^,  or  ^  of 
the  principal,  equals  the  interest.  For  a  principal  to 
double  itself,  it  must  gain  once  itself,  or  |g  of  itself; 
hence  it  will  require  as  many  years  as  ^  is  contained 
times  in  |g,  which  are  20. 

21.  SUG.  —  8280,  the  amount  at  8  per  cent.,  minus 
§250,  the  amount  at  5  per  cent.,  equals  §30,  the  interest 
on  the  given  principal  for  the  given  time  at  8  per  cent., 
minus  5  per  cent.,  or  3  per  cent.;  hence  the  interest  at 
one  per  cent,  is  \  of  §30,  which  is  §10,  and  5  per  cent. 
o  time.s  §10,  or  §50,  and  the  principal  is  §250  —  §50,  or 
§200.  The  time  is  found  as  in  the  first  problem  of  this 


OFTHE 

([   UNIVERSITY 

OF 


44  KEY.  [SECTION  vii. 

LESSON  X. 

13.  REM. — By  a  little  analysis  it  may  be  shown  that  a 
principal  will  gain  3  times  itself,  at  300  per  cent.,  in  one 
year,  and  hence  in  10  years,  at  y1^  of  300  per  cent.,  or 
80  per  cent. 

17.  SOL. — In  10  years  at  one  per  cent.,  y1^,  or  y1^  of 
the  principal,  equals  the  interest.  For  a  principal  to 
treble  itself  it  must  gain  twice  itself,  or  f g  of  itself; 
hence  it  will  require  as  many  times  one  per  cent,  as  y1^  is 
contained  times  in  |g,  which  are  20. 

23.  SUG.— The  difference  between  $600  and  §540, 
which  is  $60,  equals  the  interest  for  10  years  minus  7 
years,  which  is  3  years ;  hence  the  interest  for  one  year 
is  $20,  and  for  10  years  $200;  hence  the  principal  equals 
$600  —  $200  =  $400.  The  rate  per  cent,  is  found  as 
in  the  first  problem  of  this  lesson. 


SECTION    VII. 

LESSON  I. 

5.  SOL. — If  10  sheep  eat  as  much  as  1  ox,  100  sheep, 
which  are  10  times  10  sheep,  will  eat  as  much  as  10 
oxen ;  hence  12  oxen  and  100  sheep  will  eat  as  much  as 
12  oxen  plus  10  oxen,  which  are  22  oxen.  If  the  pas- 
turage of  22  oxen  cost  $44,  &c. 

8.  SOL. — If  2  men  do  as  much  as  3  boys,  6  men, 
which  are  3  times  2  men,  will  do  as  much  as  3  times  3 
boys,  which  are  9  boys;  hence  6  men  and  15  boys  will 
do  as  much  as  9  boys  plus  15  boys,  which  are  24  boys. 
If  24  boys  earn  $72,  &c. 

9.  SUG. — Three  men  for  5  days  will  earn  as  much  as 
15  men  for  one  day,  and  4  men  3  days  will  earn  as  much 
as  1 2  men  in  one  day,  &c. 

REM. — Some  arithmeticians  say,  3  men  for  5  days  is 
the  same  as  15  men  for  1  day,  &c.  This  is  not  quite 
true;  "will  earn"  &c..  is  better 


LESSON  II.]  KEY.  45 

15.  SUG. — Twelve  horses,  which  are  4  times  3  horses, 
will  do  as  much  as  4  times  5,  or  20  oxen ;  then  12  horses 
and  18  oxen  will  do  as  much  as  20  oxen  plus  18  oxen,  or 
38  oxen,  &c. 

18.  SUG. — We  find  a  horse  eats  as  much  as  2  cows,  or 
as  much  as  8  sheep ;  hence  18  cows  will  eat  as  much  as 
9  horses,  and  48  sheep  will  eat  as  much  as  6  horses,  and 
they  will  all  eat  as  much  as  21  horses.  If  the  pasturage 
of  21  horses  is  §63,  &c. 


LESSON  II. 

1.  HEM. — In  the  solution  of  such  problems,  some  say, 

Since  the  parts  are  as  4  to  6,  we  must  divide/'  &c. 
This  is  not  true;  we  could  divide  into  2  +  3,  which  are 
5  qqual  parts.  Instead  of  saying  must,  say,  if  we  divide 
into  4  -f-  6,  &c. 

10.  SOL. — One-half  equals  |  and  -1  equals  f ;  hence 
the  numbers  are  to  each  other  as  |  is  to  |,  or  as  3  to  2. 
Therefore,  if  we  divide  50  into  two  3  plus  2,  which  are 
5  equal  parts,  3  of  these  parts,  or  |  of  50,  will  be  one 
number,  &c. 

15.  SUG. — The  reciprocal  of  a  number  equals  unity 
divided  by  that  number.  The  reciprocal  of  2  is  A,  and 
the  reciprocal  of  3  is  \  ]  hence  they  pay  in  the  propor- 
tion of  ^  to  ^,  or  |  to  |,  or  3  to  2,  &c. 

18.  SUG. — This  problem  may  be  solved  in  several  dif- 
ferent ways.  We  suggest  two  of  the  best. 

First.  If  A's  is  to  B's  as  3  to  4,  4  times  A's  =  3  times 
B's,  or  B's  =  |  of  A's,  and  ^  of  B's  =  f  of  A's,  which, 
ridded  to  A's,  equals  |  of  A's,  which  equals  $2000,  from 
which  A's  and  B's  can  readily  be  found. 

Second.  If  A's  is  to  B's  as  3  to  4,  A's  is  to  ^  of  B's 
as  3  to  2 }  hence  if  we  divide  $2000  in  two  parts,  which 
are  as  3  to  2,  the  first  part  will  be  A's  and  the  second 
part  ^  of  B's;  &c. 


46  KEY.  [SECTION  vii. 

19.  SUG.— After  finding  the  sum  of  -}  of  A's  and  \  of 
B's,  we  proceed  thus  :  If  A's  is  to  B's  as  9  to  8,  |  of  A's 
is  to  ^  of  B's  as  3  to  2 ;  hence  if  we  divide  $500  into 
two  parts  which  are  to  each  other  as  3  to  2,  the  first  part 
will  be  -i  of  A's,  and  the  second  part  |  of  B's,  &c. 


LESSON  III. 

18.  SUG. — Subtract  what  A  and  B  do  in  a  day  from 
what  A,  B,  and  C  do  in  a  day,  and  we  have  what  C  does 
in  a  day ;  subtract  this  from  what  B  and  C  do  in  a  day, 
and  we  have  what  B  does  in  a  day ;  subtract  this  from 
what  A  and  B  do  in  a  day,  and*  we  have  what  A  does  in 
one  day,  from  which  we  can  find  the  time  in  which  each 
can  do  it. 

23.  SUG.— We  find  that  E  can  build  the  fence  in  18 
days ;  hence  he  can  build  |  of  it,  which  remains  after  D 
has  built  |  of  it,  in  ^  of  18  days,  or  6  days. 

24.  SOL. — If  2  men  can  plough  an  acre  in  i  of  a  day, 
they  can  plough  6  acres  in  one  day,  one  man  can  plough 
i  of  6  acres,  or  3  acres,  and  3  men  can  plough  9  acres 
in  a  day.     If  3  boys  can  plough  6  acres,  one  boy  can 
plough  2  acres,  and  2  boys  can  plough  4  acres ;  hence  3 
men  and  2  boys  can  plough  13  acres  in  a  day,  and  they 
can  plough  1  acre  in  y^  of  a  day. 

26.  SUG — A,  B,  and  C  did  -J  of  it;  hence  there  re- 
mained |  of  it.     A  and  B  can  do  the  whole  in  9  days; 
hence  they  can  do  f  of  it  in  f  of  9  days,  or  6  days. 

27.  SUG. — Sallie  can  make  it  in  3  days,  and  Euretta 
in  4  days ;  hence  the  three  will  make  i  -f-  \  -|~  i  —  792 
of  it  in  a  day,  and  after  working  f  of  a  day  there  re- 
mained A  of  it  to  be  made.     Marie  and  Sallie  can  make 

-(-  A  =  ^  of  it  in  one  day;  hence  it  will  require  one 
,ay  for  Marie  and  Sallie  to  finish  it 


LESSON  V.]  KEY.  47 


LESSON  IV. 

9.  SOL.— If  |  of  C's  age  equals  f  of  B's,  -f  6,  |  of 
C's  equals  §  of  C's,  +  2,  and  f  of  C's  equals  f  of  C's,  -f  8, 
which,  added  to  -|  of  C's,  equals  *g  of  C's,  -f-  8,  which 
equals  34,  &c. 

11.  SuG.: — We  find  that  f  of  Fanny's  number  equals 
Jg5  of  Sallie's  number;  then  !g5  of  Sallie's  number,  minus 
|  of  Sallie's  number,  which  is  J  of  Sallied  number, 
equals  14,  &c. 

15.  SUG. — We  find  that  3  times  what  A  builds  equals 
what  they  both  build,  or  |  of  a  boat ;  then  once  what  A 
builds  equals  -l  of  i  of  a  boat,  or  y1^  of  the  boat,  and 
twice  what  A  builds,  or  what  B  builds,  equals  ^  of  the 
boat ;  hence  it  will  require  A  18  days,  and  B  9  days  to 
build  the  boat. 

19.  SUG. — They  can  build  £  in  one  day,  which,  divided 
into  three  parts,  which  are  as  1,  2,  and  3,  gives  -Jg,  T7g, 
and  JTJ,  the  respective  parts  each  can  do  in  one  day;  hence 
it  wilf  take  A  36  days,  B  18  days,  and  C  12  days. 

22.  SUG. — B  can  drink  i  of  it  in  a  day,  A  |  of  it, 
and  C  ^  of  it  in  a  day ;  A,  B,-  and  C  can  drink  ^  +  i  + 
|  —  |j  of  it  in  a  day ;  hence  after  drinking  -fy  of  a  day 
there  will  remain  1  of  it  for  A  and  C  to  drink.  A  and 
C  can  drink  |  -|-  |  =  |  of  it  in  a  day,  and  to  drink  ^  of 
it  it  will  take  them  |  of  a  day. 


LESSON  V. 

2.  SUG. — He  can  save  $2  each  day,  and  in  40  days  he 
can  save  $80,  hence  he  loses  §80  —  $50  =  §30 ;  each 
day  he  is  idle  he  loses  $2^,  hence  to  lose  $30  it  will  re- 
quire 12  days. 

NOTE. — Some  obtain  10  days  for  the  result,  assuming 
that  he  lost  $3  each  day  he  was  idle ;  but  this  is  incor- 
5 


48  KEY.  [SECTION  vir. 

rect,  as  may  be  seen  by  an  attempt  to  prove  the  problem. 
After  paying  his  board  for  the  whole  time,  which  we 
assume  in  the  first  part  of  the  solution,  he  can  lose,  by 
being  idle,  only  his  daily  wages. 

13.  SOL. — If  he  received  $2  a  day,  and  agreed  to 
labor  for  $60,  he  agreed  to  labor  as  many  days  as  $2  is 
contained  times  in  $60,  which  are  30,  &c. 

17.  SUG. — We  find  he  can  save  $16  in  a  week,  and  in 
10  weeks  he  can  save  $160 ;  hence  he  loses  $160  —  $144 
=  $16.  Now,  in  computing  what  he  loses,  we  have  re- 
garded his  board  as  paid  for  the  whole  time,  hence  each 
day  he  is  idle  he  loses  $4;  therefore,  to  lose  $16,  he  must 
idle  4  days,  &c. 

19.  SUG. — By  the  first  condition  A  pays  |  and  B  |r 
hence  f  of  what  B  pays  equals  what  A  pays.  By  the 
second  condition  B  pays  f ,  and  G  f ,  hence  f  of  what  C 
pays  equals  what  B  pays,  and  therefore  A  pays  f  of  f  of 
what  C  pays,  or  -|  of  what  C  pays  ]  then  |  of  what  C  pays 
minus  |  of  what  C  pays,  which  is  |  of  what  C  pays, 
equals  $500,  &c. 


LESSON  VI. 

It  will  be  seen  that  the  problems  in  the  first  part  of 
this  lesson  are  of  a  more  general  character  than  those 
usually  given  by  arithmeticians.  .  As  usually  stated,  the 
requirement  is,  when  will  one  be  twice  as  old  as  the  other, 
and  the  solution  given  is  of  a  special  nature,  and  cannot 
be  applied  to  the  general  problem ;  that  is,  when  the  re- 
quirements are,  when  will  one  be  three  times,  or  four 
times,  or  five  times,  &c.,  as  old  as  the  other.  By  the 
solution  here  given  we.  cannot  only  solve  the  general 
problem,  when  future  time  is  involved,  but  also  when 
past  time  is  considered. 

9.  SUG. — First  find  the  age  of  one,  as  John,  and  then 
proceed  as  before. 


LESSON  VII.]  KEY.  <± 

11.  Suo. — Find  the  age  of  one  of  them,  as  in  LessoL 
IV.,  and  then  proceed  as  before. 

16.  SUG. — He  paid  6  cents  for  2  oranges,  and  sold  2 
for  8  cents,  and  thus  gained  2  cents  on  2  oranges ;  hencw, 
to  gain  12  cents,  which  are  6  times  2  cents,  he  sold  6 
times  2  oranges,  or  12  oranges,  &c. 

19.  SUG. — If  3  yards  cost  $1,  1  yard  cost  $i,  and  if 
4  yards  cost  $1, 1  yard  cost  ${;  hence  2  yards  cost  $^  -f- 
$£  =  £T72  :  he  sold  8  yards  for  $3,  hence  he  sold  2  yards 
for  $T92,  and  therefore  gained  $T\,  or  #£,  on  2  yards; 
hence  to  gain  §5  he  must  sell  60  yards,  &c. 

23.  SOL. — As  often  as  he  paid  2  cents  for  1  apple  he 
paid  8  cents  for  2  oranges,  hence  1  apple  and  2  oranges 
cost  10  cents,  and  3  were  sold  for  9  cents ;  hence  he  lost 
on  1  apple  and  2  oranges  1  cent,  and  to  lose  100  cents 
he  must  sell  100  times  1  apple  and  2  oranges,  which  are 
100  apples  and  200  oranges. 


LESSON  VII. 

The  problems  in  the  first  part  of  this  lesson  are  usually 
found  by  pupils  to  be  rather  difficult ;  we  propose,  there- 
fore, to  present  a  solution  of  a  few  in  a  more  analytical 
form  than  given  in  the  text-book. 

.  2.  SOL. — 5  times  O's  age  now  =  John's  age  now,  and 
5  times  O's  age  now  -f-  8  yrs.  —  John's  age  in  8  yrs.,  and 

1  time  O's  age  now  -f-  8  yrs.  =  Oliver's  age  in  8  yrs.,  then 
5  times  O's  age  now  -(-  8  yrs.  =  3  times  (once  O's  now 

-f  8  yrs.),  or 
5  times  O's  age  now  -f  8  yrs.  =  3  times  O's  now  -f-  24 

yrs.,  hence 
5  times  O's  age  now  —  3  times  O's  age  now  =  24  yrs, 

minus  8  yrs.,  or 

2  times  O's  age  now  —  16  yrs., 

1  time  O's  age  now  =  8  yrs.,  and 

5  times  O's  now,  or  John's  age  =  5x8  yrs.  =  40  yrs. 


50  KEY.  [SECTION  VII. 

3.  SOL. — 4  M's  age  now  =  Aunt's  age  now, 
4  M's  age  now  -f-  20  yrs.  =  Aunt's  age  in  20  yrs., 

1  M's  age  now  -(-  20  years  =  Aunt's  age  in  20  years, 

then 

4  M's  age  now  -f-  20  yrs.  =  2  (M's  age  now  -f-  20  yrs.),  or 
4  M's  age  now  -f-  20  yrs.  =  2  M's  age  now  -j-  40  yrs., 

hence 
4  M's  age  now  —  2  M's  age  now  =  40  yrs.  —  20  yrs.;  or 

2  M's  age  now  —  20  yrs., 
M's  age  now  =  10  yrs., 

4  M's  age  now  =  40  yrs. 

7.  SUG. — The  difference  between  25  years  ago  and  5 
years  ago  is  20  years;  hence  7  times  Willard's  age  25 
years  ago  plus  20  years  =  his  uncle's  age  5  years  ago,  &c. 

14.  SOL. — Since  2  of  the  hound's  leaps  equal  8  of  the 
hare's,  for  the  hare  to  run  as  fast  as  the  hound,  it  must 
take  8  leaps  while  the  hound  takes  2  leaps,  but,  by  the 
problem,  the  hare  takes  only  4  while  the  hound  takes  2, 
therefore  the  hound  gains  8  —  4,  or  4  of  the  hare's  leaps 
in  taking  2  leaps ;  hence  to  gain  10  leaps  he  must  take 
^  of  2,  or  ^  of  a  leap,  and  to  gain  30  leaps,  the  distance 
the  hare  is  ahead,  he  must  take  30  times  \  of  a  leap, 
which  are  15  leaps. 

16.  SUG. — We  find  that  the  thief  loses  2  steps  in 
taking  6  steps;  hence  to  lose  20  steps,  which  are  10 
times  2  steps,  he  must  take  10  times  6  steps,  or  60  steps. 

18.  SUG. — We  find  that  when  B  takes  6  steps  A  will 
take  3  steps,  and  since  6  of  B's  =  4  of  A's,  A  will  lose 
1  step  every  time  he  takes  3  steps  and  B  takes  6 ;  hence 
to  lose  10  steps,  the  distance  he  was  ahead,  A  will  take 
10  times  3;  or  30  steps,  and  B  will  take  10  times  6,  or 
60  steps. 

19.  SUG. — We  find  that  B,  in  taking  2  steps,  gains  2 
of  C's  steps ;  hence  in  taking  30  steps  he  will  gain  30  of 
C's  steps,  and  therefore  C  was  30  steps  ahead  when  they 
started. 


LESSON  VIII.]  KEY.  51 

22.  SUG.— We  find  that  6  of  B's  =  3  of  A's ;  hence 
when  B  takes  9  steps  A  goes  a  distance  equal  to  6  of  B's 
steps,  from  which  we  can  find  the  distance  in  B's  steps 
each  goes  before  they  meet,  and  reducing  the  distance  A 
goes  in  B's  steps  to  A's  steps,  we  have  the-f  number  of 
'teps  each  takes  before  they  are  together,  x 


LESSON  VIII. 

*• 

The  "  time  problems"  in  this  lesson  are  «usu#lly  re- 
garded by  pupils  as  being  among  the  most  difficult  in  the 
book.  If  the  time  be  represented  by  a  line,  ^fceiie  may 
be  made  as  simple  as  the  problems  in  Lesson1  JII*, Section 
V.  It  will  be  well  for  the  teacher  to  solve  a  f$w  pf..them 
by  the  method  suggested  below,  and  also  require  the  class 
to  indicate  them  upon  the  blackboard. 

2.  SUG. — The  line  M  N  represents  T 

the  time  from  midnight  to  noon,  and  M  '  N 

M  T  present  time.  Then  f  of  M  T  =  T  N,  which,  added 
to  |  of  M  T  =  |  of  M  T,  which  equals  M  N,  or  12 
hours,  and  |  of  M  T  =  3^6  =  7^  hours;  hence  it  is  7 
o'clock  and  12  minutes  A.  M. 

4.  SUG. — In  this  problem  M  N  rep-  N 

resents  the   time   from   midnight   to  M  *" T 

noon,  M  T  the  time  past  midnight,  and  N  T  the  time 
past  noon.  Then,  ^  of  M  T  =  N  T,  which,  subtracted 
from  §  M  T  ==  |  of  M  T,  which  equals  M  N,  or  12  hours ; 
hence  ^  of  M  T  =  3  hours,  which  equals  N  T;  hence, 
it  is  3  o'clock,  P.  M. 

11.  SUG. — In  this  problem  M'  M  T 

represents  the  time  from  midnight  to  M 
midnight,  and  M'  T,  present  time.     Then,  f  of  T  M  = 
M'  T,  which,  added  to  f  of  T  M  =  f  of  T  M,  which 
equals  M'  M,  or  24  hours  •  hence  f  of  T  M,  which  is 
M'  T  =  9  hours :  hence  the  time  is  9  o'clock,  A.  M. 
5* 


52  KEY.  [SECTION  VII, 

16.  SUG. — At  4  o'clock  the  hour  and  minute  hands 
are  .4  spaces  apart,  and,  as  in  problem  14,  to  gain  one 
space  the  minute  hand  goes  -|f  of  a  space,  to  gain  4 
4  spaces  it  must  go  4  X  if  =  f  f ?  or  4T4T  spaces ;  to  go 
one  space  requires  5  minutes,  to  go  4T4T  spaces  will  re- 
quire 4T4r  X  5  minutes  =  21T9T  minutes. 

17.  SUG. — We  find  the  time  past  9  o'clock  —  f  of  the 
time  to  midnight;  then  J  -|-  |,  or  |  of  the  time  to  mid- 
night, equals  the  time  from  9  o'clock  till  midnight,  which 
is  15  hours,  &c. 

28.  SUG. — We  find  that  the  time  past  noon  equals  | 
of  the  time  to  midnight  minus  £  of  an  hour,  hence  y  of 
the  time  to  midnight  minus  £  of  an  hour  equals  12  hours, 
and  y  of  the  time  to  midnight  will  equal  6-6  hours, 
hence  f  of  the  time  to  midnight  equals  6  hours ;  there- 
fore it  is  6  o'clock,  P.  M. 

29.  SUG. — We  find  the  two  parts  to  be  40  and  50, 
then  after  cutting  a  piece  from  the  longer,  1  of  40,  or  8 
feet  =  %  of  the  remainder,  hence  the  remainder  is  32 
feet ;  hence  there  must  be  50  —  32,  or  18  feet  cut  from 
the  longer. 


^ 
1 


LESSON  IX, 

6.  SOL.  —  If  8  men  pay  for  the  coach,  each  man  pays 
of  the  price,  but  when  they  take  in  4  persons  there  are 
2  in  all,  and  each  pays  -J^  of  the  price;  hence  -J  of  the 
price  minus  -^  of  the  price,  which  is  ^  of  the  price, 
equals  $f  ,  and  ||  of  the  price  equals  24  times  8|,  or  $18. 

6.  SOL.  2d.  —  If  the  expense  of  one  man  is  diminished 
$|,  the  expense  of  8  men  is  diminished  8  times  $|,  or 
$6  ;  hence  4  men  pay  $6,  and  1  man  pays  |  of  $6,  or  $|, 
and  8  -f-  4;  or  12  men;  would  pay  12  times  |,  or  $18. 


LESSON  X.]  KEY.  53 

9.  SOL. — If  3  times  the  square  of  his  age  is  75  years, 
the  square  of  his  age  equals  J  of  75  years,  or  25  years, 
and  his  age  equals  the  square  root  of  25  years,  which  is 
5  years. 

12.  SOL. — The  square  of  twice  a  number  (2  No.)  * 
equals  4  times  the  square  of  the  number,  which  equals 
256 ;  hence  the  square  of  the  number  equals  |  of  256, 
or  64,  and  the  number  equals  the  square  root  of  64,  or  8. 

13.  SUG. — If  the  square  of  §  of  the  number  equals 
100,  §  of  the  number  equals  the  square  root  of  100, 
which  is  10,  &c. 

16.  SUG. — The  square  of  ^  the  number  equals  {  of 
the  square  of  the  number,  and  j  of  this  is  T3g  of  the 
square  of  the  number,  &c. 

19.  SUG. — The  cube  of  f  of  the  number  equals  ^  of 
the  cube  of  the  number  ;  hence  f>>  or  £f ,  of  the  cube 
of  the  number,  minus  ^  of  the  cube,  which  is  ^  of  the 
cube  of  the  number,  equals  10,  &c. 

21.  SUG. — |  of  the  number  multiplied  by  |  of  the 
number  equals  -J  of  the  square  of  the  number,  and  the 
square  of  ^  of  the  number  equals  \  of  the  square  of  the 
number,  &c. 


LESSON  X. 

The  first  problem  in  this  lesson,  and  others  of  the  same 
class,  may  be  solved  by  the  following  method,  which  will 
be  preferred  by  some  to  the  method  given  in  the  text- 
book. 

1.  SOL. — When  for  the  mixture  I  use  1  Ib.  worth  6s. 
I  will  gain  3s.,  and  when  I  use  1  Ib.  worth  lls.  I  lose 
2s.;  hence  to  lose  Is.  I  must  use  |  Ib.,  and  to  lose  3s., 
what  I  gained  on  the  other,  I  must  use  3  times  £,  or  | 
Ib.;  hence  I  use  3  Ibs.  of  the  first  as  often  as  |  Ib.  of 
the  second,  or,  in  whole  numbers,  2  Ibs.  of  the  first  as 
often  as  31bs.  of  the  second. 


54  KEY.  [SECTION  vir 

5.  SUG. — By  the  first  method  I  find  that  I  must  use, 
for  the  mixture,  3  Ibs.  at  6  cents  as  often  as  2  Ibs.  at  11 
jents;  hence  as  often  as  I  use  4  Ibs.  at  11  cents,  I  must 
use  2  times  3  Ibs.,  or  6  Ibs.,  at  6  cents. 

5.  SUG. — By  the  second  method  I  find  that  on  1  Ib.  at 

1 1  cents,  I  lose  3  cents,  and  on  4  Ibs.  I  lose  12  cents ; 
ilso,  oh  1  Ib.  at  6  cents,  I  gain  2  cents;  hence,  to  gain 

12  cents,  what  I  lost  on  4  Ibs.  at  11  cents,  I  must  use 
as  many  Ibs.  as  2  is  contained  times  in  12,  or  6  Ibs.  at  6 
cents. 

10.  SUG. — By  comparing  I  find  I  must  take  2  Ibs.  at 
6  cents  as  often  as  3  Ibs.  at  11  cents;  also,  that  I  must 
take  1  Ib.  at  7  cents  as  often  as  1  Ib.  at  11  cents;  hence 
when  I  take  2  Ibs.  at  6  cents,  and  1  Ib.  at  7  cents,  I 
must  take  2  -j-  1,  or  3  Ibs.,  at  11  cents. 

REM. — By  the  second  method,  if  1  take  1  Ib.  at  6  cents 
and  1  Ib.  at  7  cents,  I  will  gain  3  -(-  2,  or  5  cents ;  if  I 
take  1  Ib.  at  11  cents  I  will  lose  2  cents;  to  lose  5  cents, 
what  I  have  gained,  I  must  take  |  Ibs.;  hence  the  pro- 
portion is  1,  1,  |,  or  2,  2,  5. 

17.  SUG. — We  see  he  travelled  10  miles  more  the 
sixth  day,  which  is  5  days  after  the  first,  than  on  the  first 
day;  hence  he  travelled  ^  of  10,  or  2  miles  more  each 
day,  after  the  first,  than  on  the  preceding  day. 

18.  SUG. — He  travelled  10  miles  more  the  last  day 
than  the  first,  hence  he  travelled  as  many  days,  after  the 
first,  as  2  is  contained  times  in  10,  or  5 ;  therefore  he 
travelled  6  days. 


LESSON  XI. 

7.  REM. — Problems  of  this  kind  may  be  solved  by 
two  different  processes,  which  we  distinguish  as  the  direct 
and  the  indirect  methods.  In  .the  direct  method  wo 
commence  at  the  beginning  of  the  problem;  in  the  indi- 


L7.SSON  XII.]  KEY.  55 

rect  method  we  commence  at  the  latter  part  of  the  pro- 
blem, and  work  toward  the  beginning.  The  second,  or 
indirect  method,  is  preferred.  We  give  both. 

7.  SOL.  1st. — After  borrowing  as  much  as  he  had  at 
first,  he  had  twice  his  money,  and  then  spending  4  cents, 
he  had  twice  his  money  minus  4  cents.  At  the  second 
store,  after  borrowing  as  much  as  he  had,  he  had  4  times 
his  money  at  first,  minus  8  cents,  and  after  spending  4 
cents,  he  had  4  times  his  money  minus  12  cents,  which 
equals  4  cents,  &c. 

7.  SOL.  2d. — Since  he  had  4  cents  remaining  before 
spending  4  cents,  at  the  second  store,  he  had  4  cents  -f- 
4  cents,  or  8  cents;  but  ^  of  this  he  borrowed,  hence  he 
had  2  of  8  cents,  or  4  cents,  when  he  left  the  first  store ; 
but  he  had  spent  4  cents  there,  hence,  before  spending 
these  4  cents,  he  had  4  -(-  4,  or  8  cents ;  but  one  half  of 
this  he  had  just  borrowed,  hence,  at  first,  he  had  ^  of  8 
cents,  or  4  cents. 

13.  SUG. — At  each  store  he  borrowed  2  cents  more 
than  he  spent;  hence  at  3  stores  he  would  borrow  3 
times  2  cents,  or  6  cents  more  than  he  spent,  and  since 
this  doubled  his  money,  he  at  first  had  6  cents. 


LESSON  XII. 

7.  SOL. — Since  A  receives  $24,  and  they  together 
receive  §60,  B  receives  §60  —  824,  or  $36.  Had  A 
mowed  twice  as  much  as  B  he  would  have  received  twice 
§36,  or  §72,  but  he  only  received  §24,  hence  for  mowing 
8  acres  he  should  receive  §72 — §24,  or  §48,  and  for 
mowing  one  acre  he  should  receive  |.  of  §48,  or  §6 ;  hence 
A,  who  received  §24,  mowed  as  many  acres  as  6  is  con- 
tained times  in  24,  which  are  4,  and  B,  who  received 
836,  mowed  as  many  acres  as  6  is  contained  times  in  36, 
whi^h  are  6. 


56  KEY.  [SECTION  vir. 

12.  SUG. — We  find  |  of  the  time  to  midnight  2  hours 
hence,  -f-  2  hours  -f-  2  hours,  equals  12  hours ;  hence,  -| 
of  the  time  to  midnight,  2  hours  hence,  equals  8  hours, 
and  ^  of  the  time  to  midnight  2  hours  hence,  which  was 
the  time  past  noon  2  hours  ago,  equals  \  of  8  hours,  or 
2  hours ;  hence  the  time  now  is  2  hours  plus  2  hours,  or 
4  hours  past  noon,  or  4  o' clock,  P.  M. 

'13.  SUG. — We  find  F's  age  nowj  -f- 1  year,  equals  E's 
age  4  years  ago,  and  F's  age  now,  -f-  4  years,  equals  P's 
age  4  years  hence ;  therefore  F's  age  now,  -|-  4  years  — 
A  of  (F's  now  -f-  4  years),  or  ^  of  F's  age  now,  -\-  2  years; 
hence  F's  now  minus  ^  of  F's  now,  which  is  ^  of  F's  age 
now,  equals  4  —  2,  or  2  years,  &c. 

17.  SUG. — After  7  jumped  from  the  first  field  into  the 
second,  there  were  14  more  in  the  second  than  in  the 
first ;  hence  3  times  the  number  in  the  first  field,  minus 
once  the  number  in  the  first  field,  which  is  twice  the 
number  in  the  first  field,  equals  14,  and  the  number  in 
the  first  field  is  7 ;  hence  7  +  7,  or  14,  was  the  number 
in  each  field  after  the  -J-  were  sold ;  therefore  14  =  f  of 
the  number  in  each  field  at  first. 

20.  SOL. — Since  B  received  $30,  and  both  received 
$50,  A  received  $50  —  $30,  or  $20.  Since  f  of  what 
A  digs,  -f-  4  rods  equals  |  of  what  B  digs,  i  of  what  A 
receives,  -)-  the  cost  of  digging  4  rods,  equals  |  of  what 
B  receives,  that  is— |  of  $20,  -f-  the  cost  of  digging  4  rods 
equals  |  of  $30,  or  $16,  -(-  the  cost  of  digging  4  rods 
equals  $20,  hence  the  cost  of  digging  4  rods  is  $20  — 
$16,  or  $4,  and  the  cost  of  digging  1  rod  is  $1 ;  therefore 
A,  who  received  $20,  dug  as  many  rods  as  1  is  contained 
times  in  20,  &c. 


LESSON  XIII J  KEY.  57 


LESSON  XIII. 

REM. — The  problems  in  this  lesson,  which  are  sirailai 
to  the  first,  may  also  be  solved  by  another  method,  which 
some  may  prefer  to  the  one  given  in  the  Arithmetic. 
The  author  prefers  the  one  given,  however,  for  several 
reasons.  It  seems  more  analytic,  less  algebraic,  and  does 
not  involve  fractional  parts  of  the  numbers  given,  as  the 
other  meth6d  does  in  many  of  the  problems. 

1.  SOL.  2d. — After  losing  15  cents  he  had  his  money 
minus  15  cents,  and  after  finding  ^  as  much  as  he  lost, 
he  had  |  of  his  money,  minus  |  of  15  cents,  which,  by 
(he  condition  of  the  problem,  equals  £  of  his  money; 
hence  |  of  his  money,  minus  i  of  his  money,  which  is  | 
rf  his  money,  equals  15  cents,  &c. 

13.  SUG. — If  3  persons  eat  5  loaves,  each  eats  |  of 
4  loaf;  hence  A  furnished  C  with  2  —  |,  or  |  of  a  loaf, 
and  B  furnished  C  with  3  — |,  or  |  of  a  loaf.  If  for 
-§  of  a  loaf  C  pays  20  cents,  for  -J  of  a  loaf,  what  A  fur- 
nishes, he  will  pay  ^  of  20  cents,  or  4  cents,  &c. 

REM. — This  problem  is  sometimes  solved  by  dividing 
the  money  that  C  pays  between  A  and  B,  in  proportion 
to  the  number  of  loaves  each  furnishes,  instead  of  divi- 
ding in  the  proportion  of  the  number  of  loaves  each  fur- 
nishes C.  Thus  they  would  divide  the  20  cents  between 
A  and  B,  in  the  proportion  of  2  to  3,  the  number  of 
loaves  which  A  and  B  furnish  respectively." 

17.  SUG. — We  find  that  A  and  B  each  eat  4  eggs,  and 
C  eats  8  eggs ;  hence  A  furnishes  C  with  6  —  4,  or  2 
eggs,  and  B  furnishes  C  with  10  —  4,  or  6  eggs.  If  G 
pays  16  cents  for  8  eggs,  for  one  egg  he  will  pay  2  cents, 
and  for  2  eggs,  the  number  that  A  furnishes,  he  will  pay 
4  cents,  and  for  6  eggs,  the  number  that  B  furnishes,  he 
will  pay  12  cents. 


58  KEY.  [SECTION  vn. 


LESSON  XIV. 

7.  SOL. — If  for  |  of  the  remainder,  minus  8,  he  re- 
ceived $4,  for  ^  of  the  remainder,  minus  2,  he  received 
81,  and  for  J  of  the  remainder,  minus  10,  he  would  re- 
ceive §5 ;  hence  12  +  10,  or  22,  are  worth  $16  —  $5,  or 
$11,  one  is  worth  $1,  and  for  $16  he  could  buy  as  many 
as  J-  is  contained  times  in  16,  which  are  32. 

11.  SOL. — If  he  sells  A  of  the  first  remainder,  minus 
8,   for  $22,   he  would  receive  for  §  of  the  remainder, 
minus  8,  twice  $22,  or  $44  ;  hence  "8  are  worth  $60  — 
$44,  or  $16,  one  is  worth  $2,  and  for  $60  he  could  buy 
30,  and  since  the  dog  had  killed  |  of  the  sheep,  30  is  -f 
of  the  number  he  had  at  first,  &c. 

12.  SUG. — We  find  if  he  sold  the  remainder  for  cost 
he  would  receive  60  dimes,  and  if  he  kept  15,  and  feold 
the  remainder  for  cost,  he  would  receive  30  dimes ;  hence 
the  15  are  worth  60  —  30,  or  30  dimes,.and  for  60  dimes 
he  could  buy  30 ;  hence  30  =  f  of  his  number  at  first. 

14.  SUG. — We  find  the  remainder  -f- 10  are  worth  80 
dimes,  and  the  remainder  — 10  are  worth  40   dimes ; 
hence  10  +  10,  or  20,  cost  80  —  40,  or  40  dimes,  one 
cost  2  dimes,  80  dimes  will  buy  40  ',  hence  40  —  10,  or 
80,  equals  \  the  number  at  first. 

15.  SOL. — If  10  Ibs.  of  the  mixture  contain  £  of  a 
pound  of  salt,  to  contain  f  of  a  pound  of  salt  will  require 
5  times  10,  or  50  Ibs.  of  the  mixture,  and  to  contain  2 
Ibs.  of  salt  will  require  2  times  50  Ibs.,  or  100  Ibs.  of  the 
mixture ;  hence  there  were  added  100  —  60  Ibs.,  or  40 
Ibs.  of  water. 

16.  SOL. — We  find  that  for  3  cows  there  is  |  of  an 
acre  of  ploughed  land,  hence  |  of  the  number  of  acres 
pastured  equals  the  number  of  acres  ploughed ;  hence  | 
of  the  number  pastured  equals  140,  |  the  number  pas- 
tured =  80,  hence  there  were  80  times  3  cows,  or  240 
cows. 


LESSON  XV.]  KEY.  .          52 


LESSON   XV. 

The  problems  'in  this  lesson,  similar  to  the  first,  are 
perhaps  rather  too  difficult  to  be  solved  mentally,  at  the 
outset;  the  pupils  may,  therefore,  solve  them  upon  the 
black-board  for  a  few  times,  if  desirable,  before  being 
required  to  give  a  mental  solution. 

5.  SUG. — We  find  that  f  of  the  daughter's  share 
equals  a  certain  amount,  and  f  of  her  share  equals  |  of 
this  amount;  also,  that  j  of  the  son's  share  equals  2 
times  this  amount,  and  J  of  the  son's  share  equals  f  of 
this  amount;  hence  the  shares  are  as  f  to  -§,  or  as  4f  to 
|J,  or  as  15  to  40;  hence  the  daughter's  share  is  §1500, 
and  the  son's  share  is  $4000. 

7.  SOL. — After  spending  ^  of  his  money  -f  $i>  there 
remained  ^  of  his  money  —  §i ;  then  after  spending  ^ 
of  this  plus  §^,  there  remained  ^  of  (-J-  his  money  —  $ ^) 
—  §1,  which  is  A  of  his  money  —  $  J  —  $^,  which  equals 
§3  ;  hence  ^  of  his  money  equals  $3|,  and  |  of  his  money 
equals  §15. 

HEM. — Some  solve  such  problems  by  finding  what  was 
given  away  each  time,  and  then  subtract  this  from  what 
he  had  to  find  the  remainder.  Thus,  suppose  |  was 
given  away,  they  would  find  |  of  what  they  had,  and 
subtract  this  from  what  they  have.  The  better  way  is  to 
reason  thus :  if  they  gave  f  of  what  they  have  away, 
there  would  remain  i  of  what  they  have,  &c.  This  is 
just  one-half  as  long  as  the  other  method,  and  is  equally 
'  clear. 

13.  SUG. — A's  money  was  on  interest  6  years,  hence 
|  of  A's  principal  equals  his  amount,  and  B's  money  wag 
on  interest  2  years,  hence  §  of  B's  principal  equals  his 
amount ;  but  |  of  A's  amount  equals  J  of  B's  amount, 
hence  A's  amount  equals  |  of  B's  amount,  hence  f  of 
A's  principal  equals  J  of  B's  amount,  and  f  of  A's  prin- 
cipal equals  |  of  B's  amount,  and  since  f  of  B's  principaJ 

6 


60  KEY.  [SECTION  vn. 

equals  B's  amount,  |  of  B's  principal  equals  |  of  B's 
amount,  hence  A's  principal  is  to  B's  as  |  to  f ,  or  as  ^ 
to  f£,  or  as  9  to  20;  hence  A's  share  is  $90  and  B's 
share  is  $200. 

14.  SOL. — By  a  condition  of  the  problem  7 -times  the 
value  of  the  silver  watch  equals  the  value  of  the  gold 
watch  and  chain,  but* the  gold  watch  equals  3  times  the 
value  of  the  silver  watch  and  chain,  therefore  7  times  the 
value  of  the  silver  watch  equals  3  times  the  value  of  the 
silver  watch,  plus  4  times  the  value  of  the  chain ;  hence 
7  times  —  3  times,  or  4  times  the  value  of  the  silver 
watch  equals  4  times  the  value  of  the  chain,  therefore 
once  the  value  of  the  silver  watch  equals  the  value  of 
the  chain,  or  $20,  &c. 


LESSON  XVI. 

23.  SOL. — He  bought  the  goods  for  100  per  cent., 
minus  20  per  cent.,  which  is  80  per  cent,  of  par  value, 
and  sold  them  for  100  -f-  20,  or  120  per  cent,  of  par 
value;  hence  he  gained  120  —  80,  or  40  per  cent,  on  80 
per  cent.,  and  therefore  he  gained  ^  of  the  cost,  which 
equals  $90,  and  |  of  the  cost  equals  twice  $90,  or  $180. 

BEM. — This  problem  is  frequently  solved  incorrectly, 
by  regarding  $90  as  40  per  cent.,  or  |  of  the  money 
invested,  from  which  the  money  invested  is  found  to  be 
8225. 

25.  SuG. — We  find  he  sold  it  for  |  of  120  per  cent., 
or  40  per  cent. ;  hence  he  lost  100  —  40,  or  60  per  cent. 

29.  SUG.— I  sell  the  hay  for  $12  a  ton;  hence  $12 
was  |  of  the  price  asked,  and  the  price  asked  was  there- 
fore $15  a  ton. 

33.  SUG. — I  retail  for  150  per  cent.,  and  sell  at  whole- 
sale for  |  less,  or  |  of  150  per  cent.,  or  112^  per  cent.; 
hence  I  gain  12^  per  cent. 


LESSON  XVII.]  KEY.  61 

38.  SUG. — There  remained  80  per  cent.,  and  this  was 
sold  for  J  of  80  per  cent,  of  the  cost  of  the  whole,  which 
is  112  per  cent;  hence  the  gain  was  12  per  cent. 

45.  SUG. — Since  the  difference  between  the  fortunes 
is  once  C's  fortune,  the  difference  between  the  amounts 
is  once  C's  amount,  which  is  |  of  C's  principal ;  hence  * 
of  C's  fortune  equals  $330,  &c. 

45.  Another  method,  which  is  more  general,  is  as  fol- 
lows :  We  find  that  |  of  C's  principal  equals  C's  amount, 
and  |  of  C's  principal  equals  B's  amount;  then  §  of  C's 
principal,  minus  |  of  C's  principal;  which  is  |  of  C's 
principal,  equals  §330,  &c. 


LESSON  XVII. 

8.  SUG. — After  working  9  days,  |  of  the  time  re- 
quired, A  and  B  could  complete  it  in  6  days,  but  A,  B, 
and  C  completed  it  in  4  days;  hence  C  could  do  \  —  £, 
or  TV  of  the  remainder  in  one  day,  and  he  will  do  the 
remainder  in  12  days;  and  if  he  could  do  J  of  the  work 
in  12  days,  he  could  do  the  whole  of  it  in  30  days. 

14.  HEM. — This  problem  means  how  much  did  C  gain 
on  I)  when  C  had  run  40  rods. 

24.  SUG. — There  remained  6  melons,  and  they  were 
sold  for  |  of  36  cents,  or  48  cents;  hence  they  were  sold 
for  8  cents  apiece. 

27.  SUG. — He  must  sell  80  per  cent,  for  120  per  cent. ; 
hence  on  80  per  cent,  he  gains  40  per  cent.,  or  the  gain 
is  50  per  cent. 

32.  SUG. — |  of  the  cost  of  the  first  horse  equals  what 
was  received  for  it,  and  |  of  the  cost  of  the  second  horse 
equals  what  was  received  for  it ;  and  since  the  second 
horse  cost  |  as  much  as  the  first,  |  of  the  cost  of  the 


62  ,  KEY.  [SECTION  VII. 

eocond  horse  equals  f  of  the  cost  of  the'  first,  which, 
added  to  |  of  the  cost  of  the  first,  equals  J  of  the  cost 
of  the  first,  which  equals  $210,  &c. 

33.  SUG. — We  find  that  £  of  the  cost  of  the  horse 
equals  what  it  was  sold  for,  and  |  of  the  cost  of  the 
carriage  equals  what  it  was  sold  for,  but  |  of  the  cost  of 
the  horse  equals  |  of  the  cost  of  the  carriage ;  then  |  of 
the  cost  of  the  carriage,  plus  |  of  the  cost  of  the  car- 
riage, which  is  ||  of  the  cost  of  the  carriage,  equals 
$230,  &c. 


CONTRACTIONS 

AND 

ABBREVIATED  PROCESSES. 


THE  improved  methods  of  teaching  Arithmetic  at  the 
present  day,  cultivate  the  reasoning  faculties,  but  neg- 
lect, to  some  extent,  to  give  that  readiness  in  mechanical 
computation  that  was  secured  by  the  old  methods. 
Pupils  seem  to  acquire  a  distaste  for  the  mechanical 
operations  of  Addition,  Multiplication,  &c.,  and  teachers 
are  somewhat  to  blame  for  it. 

Arithmetic  consists  of  a  reasoning  and  a  mechanical 
part,  and  both  should  be  thoroughly  taught,  since  both 
are  necessary  to  make  the  accomplished  arithmetician. 
The  utility  of  readiness  in  the  mechanical  operations  is 
felt  in  the  extraction  of  roots,  the  computation  of  loga- 
rithms, and  logarithmic  sines,  &c.,  and  in  the  calculation 
of  eclipses  and  the  occultation  of  stars.  We  propose  to 
suggest  a  number  of  exercises  for  the  use  of  pupils  in 
acquiring  readiness  in  the  mechanical  processes. 


ADDITION  AND  SUBTRACTION. 

Quite  a  large  number  of  methods  is  suggested  in  the 
Normal  Primary   Arithmetic.     We    here   give   a   few 
more,  both  for  Mental  and  Written  Arithmetic. 
6*  (63) 


64  CONTRACTIONS    AND 

1.  Kequire  pupils  to  add  mentally  such  problems  as 
tlie  following : 

25  +  37;  82  +  69;  76  +  57;  124  +  367;  256  + 
385;  467  +  583;  73  +  85  +  97;  125  +  327  +  562. 

REMARK. — With  a  little  practice  pupils  will  perform 
such  problems  with  great  facility  and  accuracy,  and  such 
ability  will  be  of  much  advantage  to  them  in  the  practical 
affairs  of  life.  Let  many  such  problems  be  given  by  the 
teacher,  increasing  in  difficulty  with  the  advancement  of 
the  class. 

2.  The  following  exercises  will  also  be  valuable.    Find 
the  value  of 

5X4  +  6X7;  6x9 +  7x8;  3x9  +  4x7; 
4X6  +  3X9  +  8X5;  8X3  +  9X6  +  5X9. 

REMARK. — The  teacher  may  exercise  his  pupils  for 
quite  a  long  time  on  problems  of  this  character,  making 
them  more  difficult  as  the  class  becomes  prepared  for 
them.  Such  exercises  are  important,  also,  since  they 
prepare  the  student  for  an  abbreviated  method  of  Multi- 
plication. 

3.  The  teacher  may  arrange  columns  of  figures  upon 
the  board,  as  is   suggested   in    the    "  Normal  Primary 
Arithmetic,7'  page  31,  and  assign  with  these  columns 
problems  similar  to  the  above. 

4.  In  Written  Arithmetic  require  the  pupils  to  add, 
not  only  one  column  at  a  time,  but  two,  and  even  tliree, 
at  the  same  time.     With  practice,  two  columns  may  be 
added   at   the    same   time    very   easily.      Many    skilful 
accountants  run  up  two  columns  of  their  books  with 
much  ease. 

Many  of  the  exercises  which  we  have  suggested  for 
addition  can  also  be  used  in  subtraction.  Let  exercises 
in  addition  and  subtraction  be  combined. 

The  above  exercises,  and  others  that  the  ingenious 
teacher  may  devise,  will  be  found  very  valuable  in  check- 


ABBREVIATED    PROCESSES.  65 

ing  the  tendency  of  our  improved  methods  of  instruction, 
to  ignore  the  utility  of  the  mechanical  processes  of 
Arithmetic. 


MULTIPLICATION. 

The  following  exercises,  in  Multiplication,  are  pre- 
sented for  the  purpose  of  acquiring  readiness  in  the  me- 
chanical process  of  multiplying.  Some  of  the  following 
problems  may  be  solved  entirely  mentally,  others  are 
adapted  to  slate  and  black-board  exercises. 

1.  To  multiply  when  the  multiplicand  and  multiplier 
consists  of  several  figures. 

1.  What  is  the  product  of  46  X  5?  289  X  6?  572 
X8?  896X7?  785X5?  469x9?  25x34?   87 
X53?  98  X37?  123  X234? 

Many  problems  in  Multiplication  may  be  solved  by 
deriving  a  law  by  which  the  product  arises.  The  fol- 
lowing are  presented  as  an  illustration. 

2.  To  multiply  two  numbers  between  10  and  20. 
METHOD. — Arrange,  from  right  to  left,  the  right  hand 

figure  of  the  product  of  the  units,  the  left  hand  figure  of 
this  product,  plus  the  sum  of  the  units,  and  then  the  pro- 
duct of  the  tens. 

DEM. — This  law  can  easily  be  derived  by  multiplying  together 

two  numbers,  as  16  by  18 ;  thus,  16  X  18  =  1  —  (8  +  6  +  4)  —  8 
r^288. 

What  is  the  value  of 

12X13?  15X14?  16X17?  13x18?  17x19? 
15  X  18  ? 

3.  To  multiply  two  numbers  of  two  places  when  the 
unit  figure  is  1  in  each. 

METHOD. — Arrange,  from  right  to  left,  the  unit  figure, 
the  sum  of  the  tens,  and  the  product  of  the  tens. 


66  CONTRACTIONS    AND 

DEM. — This  law    can    be    readily    derived    by  multiplying 

together    two    numbers,    as    41    and    71 ;    thus,    41  X  71  =» 

7X4  —  (7-J-4)  — 1  =  2911. 

What  is  the  value  of 

21X31?  21X51?   51X41?  71x81?  91x21? 
81  X91? 

4.  To  multiply  two  numbers  of  two  places   each,  in 
which  the  unit  figures  are  alike,  and  the  sum  of  the  tens 
is  10. 

METHOD. —  Write  from  right  to  left  the  product  of  the 
units,  and  the  product  of  the  tens  increased  l)y  the  unit 
figure  of  the  numbers. 

DEM. — This  can  be  readily  shown  by  multiplying  together 
two  numbers,  as  36  and  76 ;  thus,  76  X  36  =  (7  X  3)»— 
(8  X  6  +  7  X  6),- (6  X  6)  =  (7  X  3),- (10  X  6),- 6  X  6 

What  is  the  value  of 

42X62?    37X77?    85x25?   98x18?   79x39? 
39  X  79  ? 

5.  To  multiply  two  numbers  of  two  places,  when  the 
sum  of  the  units  is  10,  and  the  difference  of  the  tens  is  1. 

METHOD.—  Write,  from  right  to  left,  100,  minus  the 
square  of  the  units  of  the  larger  number,  and  to  the  right 
of  this  the  square  of  the  tens  of  the  larger  number  dimin- 
ished Ity  1. 

DEM. — This  may  be  readily  shown  by  multiplying  together 
two  numbers,  as  57  and  63,  thus  : 

57  X  63  =  (6)2~  1  — (100  —  (3)2)  =  3591. 

What  is  the  value  of 

46X54?   87X43?   52x68?   79x61?   86x94? 
71  X  89  ? 

6.  To  multiply  any  integer  -j-  £  by  itself. 


ABBREVIATED   PROCESSES.  67 

METHOD. — Multiply  the  whole  number  ly  the  next 
larger  whole  number,  and  to  the  product  add  i. 

DEM. — This  may  be  proved  by  multiplying  any  two  such 
mixed  numbers  together,  as  7J  and  7J. 

7}  X  7J  =  (7  -f  J)  (7  +  })  =  (7  X  7  +  (J  X  7  +  J  X  7 )  +  i 
Xi  =  7X7-f-lX7  +  i  =  (7Xl)7  +  i  =  «X7  +  J, 
which  equals  56J-. 

What  is  the  value  of 

81X8A?  9±X9A?  Hi  X  H£?  14-JX14A?  15i 
Xloif  mxl7|?  25^X251?  75- X  75=*  7| 
tens  X  7£  tens  ?  95  X  95  ? 

7.  To  multiply  two  mixed  numbers  when  the  integers 
are  th'e  same,  and  the  sum  of  the  fractions  equals  1. 

METHOD. — Multiply  the  whole  number  by  the  next 
larger  whole  number,  and  to  the  product  add  the  product 
of  the  fractions. 

DEM. — This  may  be  readily  derived  by  multiplying  together 
two  such  numbers   as   5|  and  54;  thus,  53  X  5i  =  (5  -{-  -§) 
(•5  +  4)  -5  X  5  +  (|  X  5  +  4  X  5)  +  3  X  4  =6  X  5  +  1  . 
X  &  +  |  X  4  -  6  X  5  +  if  =  30||. 

What  is  the  value  of 

<5§X6f?  9|X9|?  15/TX15T4T?  20^X20^? 
2'V0X25TV  30  AX  30  A?  40&  X  40ft?  50^ 
X  50|| ? 

8.  To  multiply  two  mixed  numbers  when  the  difference 
of  the  integers  is  a  unit,  and  the  sum  of  the  fractions  is 
a  unit. 

METHOD. — First  square  the  larger  number,  and  di- 
minish this  by  \y  secondly )  square  the  fraction  of  the 
lirger  number,  subtract  this  from  1,  and  unite  the  result 
with  the  former  result. 

DEM. — This  law  may  be  derived  by  a  process  similar  to  those 
tlready  explained. 

What  is  the  value  of 
•HX41?   5'JX6f?   7fX6|?   8|X9|?  9|XlOJ? 


68  CONTRACTIONS   AND 

12jxl3f?    15$xl6J?     17§X18|?     19|  X  18  J 
X  20i  X  21|  ? 

9.  To  find  the  product  of  any  two  mixed   numbers, 
whose  fractional  parts  are  halves. 

METHOD.  —  Take  the  product  of  the  integers,  increase 
this  by  ^  of  their  sum,  and  by  |. 

DEM.  —  This  law  can  be  readily  obtained  by  a  process  similar 
to  those  already  given. 

What  is  the  value  of 
S-i  X  5|  ?     6|  X  2^  ?     7^  X  54  ?     6J  X  9±  ?     10:! 


10.  To  find  the  square  of  a  mixed  number,  whose  frac- 
tional part  is  l. 

METHOD.  —  First,  when  the  integer  is  EVEN,  square  the 
integer,  add  ^  of  itself  and  the  square  of  1  . 

Secondly,  when  the  integer  is  ODD,  square  the  whole 
number,  add  1  of  the  next  smaller  number,  and  also  T9g. 

DEM.  —  This  law  may  be  derived  by  a  process  similar  to  those 
already  given. 

What  is  the  square  of 

4'?  61?  8'?  10'?  121?  14|?  161?  181? 
204?  24'?  51?  74?  94?  1U?  134?  154? 
17|?  19|?  211?  251? 

11.  To  find   the  square  of  a  mixed   number,  whose 
fractional  part  is  |. 

METHOD.  —  First,  when  the  number  is  EVEN,  square 
the  whole  number,  increase  this  by  |  of  the  whole  num- 
ber, and  by  the  square  of  \. 

Secondly,  when  the  number  is  ODD,  square  the  whole 
number,  increase  this  by  |  of  the  next  smaller  number, 
also  by  2  and  by  T3g. 

What  is  the  square  of 

4|?  6|?  8|?  lOf?  12|?  14|?  18|?  20|? 
24|?  30|?  7|?  9|?  11|?  15|?'l7|?  19|  ? 
25|?  27|?  33|?  41|? 


ABBREVIATED   PROCESSES.  69 

12.  To  square  any  number  whose  unit  figure  is  5. 

METHOD. — Multiply  the  part  preceding  the  units  by 
itself,  increased  by  a  unit,  and  prefix  the  product  to  25 ; 
thus,  in  squaring  65,  we  have  6  X  7,  or  42,  which,  pre- 
fixed to  25,  gives  4225. 

DEM. — This  law  may  be  also  derived  by  a  process  similar  to 
those  already  given. 

What  is  the  square  of 

25?  35?  45?  55?  75?  85?  95?  105?  115? 
125?  135?  145?  155?  165?  175?  185?  195? 
205  ?  225  ? 

13.  To  find  the  product  of  any  two  numbers  whose 
unit  figures  are  5. 

METHOD. — Take  the  product  of  the  figures  preceding 
the  5  in  each  number,  increase  this  by  ^  of  tJie  sum  of 
these  figures,  and  prefix  the  result  to  25. 

DEM. — This  method  is  also  derived  by  a  process  similar  to 
those  already  explained. 

REMARK. — If  the  sum  of  the  figures  preceding  the  5 
is  ODD,  when  we  take  i  of  it,  the  £,  or  5  tens,  which  re- 
mains, must  be  added  to  the  figure  2  of  the  25,  or  we 
may  take  ^  of  the  next  smaller  number,  and  use  75  as 
the  suffix. 

What  is  the  value  of 

25X45?  55X75?  75x95?  65x95?  35x85? 
85X45?  155X35?  165x45?  185x65?  175 
X  65  ?  225  X  105  ? 

14.  To  find  the  square  of  any  number  ending  in  25. 

METHOD. — Regard  the  25  as  |,  and  then  proceed  as 
in  the  case  of  squaring  a  mixed  number,  whose  fraction  al 
part  is  ^,  and  in  the  result  reduce  the  number  of  sixteenths, 
which  is  j1^  of  a  number  of  thousands,  to  tens  and  units. 

What  is  the  square  of 

225?  425?  625?  825?  1025?  1225?  2025? 
125?  325?  525?  725?  925?  1325?  2125? 


70  CONTRACTIONS    AND 

15.  To  multiply  any  numbers  in  which  the  sum  of  the 
units  and  tens  equals  100. 

METHOD. —  This  is  similar  to  multiplying  two  mixed 
numbers j  in  which  the  sum  of  the  fractions  equals  a  unit. 
We  regard  the  number  expressed  by  the  units  and  tens  as 
so  many  hundredths,  and  proceed  as  in  a  previous  case. 

EEMARK. — This  method  is  particularly  adapted  to  two 
numbers,  one  of  which  ends  in  25,  and  the  other  in  75, 
since  these  are  respectively  equal  to  |  and  |. 

What  is  the  value  of 

125X175?  225X475?  675x825?  975x725? 
1025X1275?  125X275?  375x^25?  325x875? 
825  X575?  1375  X  1425? 

16.  To  find  the  square  of  any  number  ending  in  75. 

METHOD. —  We  regard  the  75  as  |,  and  then  proceed- 
as  in  the  case  of  squaring  a  mixed  number ,  whose  frac- 
tional part  is  |. 

17.  To  multiply  by  aliquot  parts  of  10, 100, 100),  &c., 
as2|-;  31;  5;  12';  16§ ;  25;  331;  50;  125;   166| ; 
250;  333 A;  833 J. 

18.  To  multiply  any  number  by  2^-. 

METHOD. —  We  annex  a  cipher  to  the  right  of  the  mul- 
tiplicand, and  divide  by  4. 

DEM. — 21  =  i-O,  hence  2|  times  a  number  equals  j.  of  10 
times  the  number. 

What  is  the  value  of 

4X21?  16X21?  28X2-I?  156  X  2|  ?  258 
X2'?~  364X21? 

19.  To  multiply  any  number  by  31. 
METHOD. —  We  annex  a  cipher  and  divide  by  3. 

DEM. — This  and  the  following  methods  maybe  demonstrated 
in  a  manner  similar  to  the  preceding. 


ABBREVIATED   PROCESSES.  71 

What  is  the  value  of 
9  X  31  ?     27  X  3^  ?     837  X  3^  ?     757  X  3j  ?     586 

20.  To  multiply  by  5. 

METHOD. — Annex  one  cipher  and  divide  by  2. 

REMARK. — The  teacher  can  form  problems  for  himself, 
illustrating  the  other  cases. 

21.  To  multiply  by  121. 

METHOD. — 12^  being  J  oy  100,  we  annex  two  naugnts 
and  divide  by  8. 

22.  To  multiply  by  16f . 

METHOD. — 16|  being  |  o/*100,  we  annex  two  naughts 
and  divide  by  6. 

23.  To  multiply  by  25. 

METHOD. — 25  being  |  of  100,  we  annex  two  naughts 
and  divide  by  4. 

24.  To  multiply  by  33i. 

METHOD. — 33|  being  |  of  100,  we  annex  two  naughts 
and  divide  by  3. 

25.  To  multiply  by  50. 

METHOD. — Since  50  is  A  of  100,  ice  annex  two  naughts 
and  divide  by  2. 

26.  To  multiply  by  125,  166|,  250,  333'. 

METHOD. — Since  these  numbers  are  respectively  |,  1, 
|,  and  -i  of  1000,  we  annex  three  naughts,  and  divide 
respectively  by  8,  6,  4,  and  3. 

REMARK. — Other  methods,  somewhat  similar  to  these, 
may  be  obtained  by  the  teacher;  we  give  one  as  an  illus- 
tration. 

27.  To  multiply  by  133f 

METHOD.— Since  133  J  is  |  of  100,  we  annex  two 

7 


72  CONTRACTIONS   AND 

naughts  to  the  tt+ultiplicimd,  divide  by  3,  and  add  the 
quotient  to  the  dividend;  thus,  369  X  133^  =  36900 
;    3  +  3690tf  =  49200. 
The  work  may  be  expressed  thus : — 
#136900 


12300      . 

49200  Ans. 

What  is  the  value  of 
v  v  X  133^  ?      684  X  133|  ?      828  X  133  J  ?      576 


28.  To  multiply  by  a  number  consisting  of  nines. 

METHOD.  —  Annex  as  many  naughts  to  the  multipli- 
cand as  there  are  9's  in  the  multiplier,  and  subtract  the 
multiplicand  from  the  product. 

DEM?—  9  fimes  a  number  =  (10  —  1)  times  the  number  =  10 
times  the  number  minus  once  the  number  ;  in  the  same  way, 
99  times  a  number  equals  100  times  the  number,  minus  once 
the  number,  &c. 

What  is  the  value  of 

28X9?  215X99?  316x99?  282x99?  351 
X  999  ?  ,  462  x  99  ?  715  X  999  ?  1867  X  9999  ? 
7857  X  9999  ? 

29.  To  multiply  by  any  number  in  which  all  the  digits 
are  the  same. 

METHOD.  —  Multiply  by  a  number  consisting  of  as  many 
9's  as  there  are  similar  digits,  and  take  such  apart  of 
tlie  product  as  the  digit  of  the  multiplier  is  of  9. 

DEM.  —  Suppose  we  wish  to  multiply  by  tfome  such  number  as 
444  ;  this  number  is  1  of  999  ;  hence  4  Of  999  times  a  number, 
equals  444  times  the  number. 

REMARK.  —  This  is  convenient  when  the  similar  digits 
are  1,  3,  or  6,  in  which  case  A,  -J,  and  f  are  the  respect- 
ive parts  of  the  product  of  9's  taken. 

30.  In  the  ordinary  processes  of  multiplication,  we 


ABBREVIATED   PROCESSES.  73 

obtain  partial  products,  and  then  add  these  together  for  the 
entire  product.  With  a  little  practice,  however,  we  may 
multiply  by  a  number  consisting  of  several  figures  with- 
out writing  the  partial  products.  There  have  been  those 
who  could  multiply  by  a  number  consisting  of  10  or  12 
digits,  writing  the  result  under  the  given  numbers  with 
great  readiness.  This  is  a  very  unusual  degree  of  profi- 
ciency ;  but  almost  any  one  can  learn  to  do  the  same  with 
a  multiplier  consisting  of  from  2  to  6  places.  We  indi- 
cate the  method  by  the  following  problem  and  its  solution. 

1.  Multiply  5642  by  345. 

EXPLANATION. — 1st.  5  X  2  =  10»  we  write  °>  and  reserve  the 

1  to  "carry." 

5642  2d-  5X4  +  4X2  +  1  =  29,  write  9,  and  re- 

345         serve  the  2  to  "carry." 

3d.  5X6  +  4X4  +  3X2  +  2  =  54,  write  4, 

1946490        reserve  the  5  to  carry. 

4th.  5  X  5  +  4  X  6  +  3  X  4  +  5  =  66,  write  6, 
reserve  the  6  to  carry. 

6th.  4  X  5  +  3  X  6  +  6  =  44,  write  4,  reserve  4  to  carry. 
6th.  3  X  5  +  4  =  19  ;  write  this,  since  it  is  the  last  product. 

The  above  work,  of  course,  is  performed  mentally ;  the 
products,  and  the  sum  of  the  products,  obtained  as  we 
have  indicated.  With  practice,  a  person  may  multiply 
in  this  manner  with  much  rapidity  and  accuracy. 


DIVISION. 

Many  of  these  processes,  which  we  have  given  for 
multiplication,  may,  by  a  little  change,  be  applied  to  divi- 
sion. In  dividing  the  aliquot  parts  of  10,  100,  &c.;  we 
just  reverse  the  method  used  in  multiplication. 

To  divide  by  2^,  we  multiply  by  4  and  divide  by  10. 

To  divide  by  16§,  we  multiply  by  6  and  divide  by  100. 

To  divide  by  25,  we  multiply  by  4  and  divide  by  100. 

In  a  similar  manner,  we  divide  by  the  other  aliquot 
parts  of  10,  100,  &c. 

Other  methods  may  be  presented  by  the  teacher ;  and 
we  suggest  that  the  exercise  will  be  valuable  to  the  student. 


SOCIAL  ARITHMETIC. 


UNDER  the  head  of  Social  Arithmetic,  we  give  a  collec- 
tion of  problems,  particularly  adapted  to  a  social  circle, 
or  the  fireside  of  a  winter  evening.  The  most  of  these 
problems  are  in  the  form  of  puzzles,  and  some  of  them 
particularly  amusing.  The  majority  of  them  are  very  old, 
their  parentage  being  entirely  unknown ;  so  that  no  credit 
can  be  given  to  their  authors.  Quite  a  number  of  them, 
however,  have  not  previously  been  published. 

1.  Think  of  a  number  of  3  or  more  figures,  divide  by 
9,  and  name  the  remainder ;  erase  one  figure  of  the  num- 
ber, divide  by  9,  and  tell  me  the  remainder,  and  I  will 
tell  you  what  figure  you  erased. 

METHOD. — If  the  second  remainder  is  less  than  the  first,  the 
figure  erased  is  the  difference  between  the  remainders  ;  but  if 
the  second  remainder  is  greater  than  the  first,  the  figure  erased 
equals  9,  minus  the  difference  of  the  remainders. 

2.  Think  of  a  number,  multiply  it  by  3,  and  multiply 
it  also  by  4,  take  the  sum  of  the  squares  of  the  products, 
extract  the  square  root  of  this  sum,  divide  by  the  first 
number,  and  I  will  name  the  quotient. 

METHOD. — The  quotient  will  always  be  5.  The  same  will  be 
also  true  if  we  have  them  multiply  and  divide  by  the  same 
multiples  of  3,  4,  and  5,  as  6,  8,  10,  &c.  If  we  have  them 
divide  by  5,  it  will  give  the  number  they  commenced  with. 

3.  Think  of  a  number,  multiply  it  by  5,  also  by  12; 
square  each  product,  take  their  sum,  extract  the  square 
root,  divide  by  the  number  commenced  with,  and  I  will 
name  the  quotient. 

(74) 


SOCIAL   ARITHMETIC.  75 

METHOD. — The  quotient  is  always  13.  To  give  variety  it  is 
well  to  use  multiples  of  5,  12 ;  as  10,  24,  &c.,  and  then  the 
quotient  is  26,  &c. 

4.  Think  of  a  number  composed  of  two  unequal  digits, 
invert  the  digits,  take  the  difference  between  this  and 
the  original  number,  name  one  of  the  digits  and  I  will 
name  the  other. 

METHOD. — The  sum  of  the  digits  in  the  difference  is  always 
9 ;  hence  when  one  is  named,  the  other  equals  9  minus  the  one 
named. 

5.  Take  any  number  consisting  of  three  consecutive 
digits  and  permutate  them,  making  6  numbers,  and  take 
the  sum  of  these  numbers,  divide  by  6,  and  tell  me  the 
result,  and  I  will  tell  you  the  digits  of  the  number  taken. 

METHOD. — The  quotient  consists  of  three  equal  digits ;  the 
digits  of  the  number  taken  are,  1st,  one  of  these  equal  digits ; 
2d,  this  digit  increased  by  a  unit ;  3d,  this  digit  diminished  by 
a  unit.  The  same  principle  holds  when  the  digits  of  the  num- 
ber taken  differ  by  2,  3,  or  4.  It  is  a  very  pretty  problem  to 
prove  that  the  sum  is  always  divisible  by  9,  and  18. 

6.  Think  of  a  number  greater  than  3,  multiply  it  by  3 ; 
if  even,  divide  it  by  2 ;  if  odd,  add  1,  and  then  divide  by 
2.     Multiply  the  quotient  by  3 ;  if  even,  divide  by  2 ; 
if  odd,  add  1,  and  then  divide  by  2.     Now  divide  by  9 
and  tell  the  quotient,  without  the  remainder,  and  I  will 
tell  you  the  number  thought  of. 

METHOD. — If  even  both  times,  multiply  the  quotient  by  4; 
if  even  2d,  and  odd  1st,  multiply  by  4,  and  add  1 ;  if  even  1st, 
and  odd  2d,  multiply  by  4,  and  add  2 ;  if  odd  both  times,  mul- 
tiply by  4,  and  add  3. 

6.  Take  any  number,  divide  it  by  9,  and  name  the 
remainder.  Multiply  the  number  by  some  number  which 
I  name,  and  divide  this  product  by  9,  and  I  will  name 
the  remainder. 

METHOD. — To  tell  the  remainder,  I  multiply  the  first  re- 
mainder by  the  number  by  which  I  told  them  to  multiply  the 
given  number,  and  divide  this  product  by  9.  The  remainder 
is  the  second  number  that  they  obtained. 

7* 


76  SOCIAL   ARITHMETIC. 

7.  A  and  B  have  an  8  gallon  cask  full  of  wine,  which 
they  wish  to  divide  into  two  equal  parts,  and  the  only 
measures  they  have  are  a  5  gallon  cask  and  a  3  gallon 
cask.     How  shall  they  make  the  division  with  these  two 
vessels  ? 

8.  Two  men  have  24  ounces  of  fluid,  which  they  wish 
to  divide  between  them  equally.     How  shall  they  effect 
the  division,  provided  they  have  only  three  vessels ;  one 
containing  5  oz.,  the  other  11  oz.,  and  the  third  13  oz.  ? 

9.  Two  men,  stopping  at  an  oyster  saloon,  laid  a  wager 
as  to  which  could  eat  the  most  oysters.     One  eat  ninety- 
nine,  and  the  other  eat  a  hundred  and  won.     How  many 
did  both  eat  ? 

REMARK. — The  "catch"  is  in  "a  hundred  and  won."  When 
this  is  repeated  it  sounds  as  if  it  meant  "one  eat  99  and  the 
other  eat  101 ;  hence  the  result  usually  given  is  200.  The 
correct  result,  of  course,  is  199. 

10.  Six  ears  of  corn  are  in  a  hollow  stump.  How  long 
will  it  take  a  squirrel  to  carry  them  all  out,  if  he  takes 
out  three  ears  a  day  ? 

REMARK. — The  "catch"  is  in  the  word  ears.  He  carries 
out  two  ears  on  his  head,  and  one  ear  of  corn  each  day ;  hence 
it  will  take  him  6  days. 

11.  A  and  B  went  to  market  with  30  pigs  each.     A 
sold  his  at  2  for  $1,  and  B  at  the  rate  of  3  for  $1,  and 
they,  together,  received  $25.     The  next  day  A  went  to 
market  alone  with  60  pigs,  and,  wishing  to  sell  at  the 
same  rate,  sold  them  at  5  for  $2,  and  received  only  $24. 
Why  should  he  not  receive  as  much  as  when  B  owned 
half  of  the  pigs  ? 

12.  In  the  bottom  of  a  well,  45  feet  deep,  there  was  a 
frog  which  commenced  travelling  towards  the  top.     In 
his  journey  he  ascended  3  feet  every  day,  but  fell  back 
2  feet  every  night.     In  how  many  days  did  he  get  out  of 
the  well  ? 

13.  A  man  having  a  fox,  a  goose,  and  some  corn, 


SOCIAL  ARITHMETIC.  77 

came  to  a  river  which  it  was  necessary  to  cross.  He 
could,  however,  take  only  one  across  at  a  time,  and  if  he 
left  the  goose  and  corn,  while  he  took  the  fox  over,  the 
goose  would  eat  the  corn ;  but  if  he  left  the  fox  and 
goose,  the  fox  would  kill  the  goose.  How  shall  he  get 
them  all  safely  over  ? 

14.  A  man  went  to  a  store  and  purchased  a  pair  of 
boots  worth  $5,  and  hands  out  a  $50  bill  to  pay  for  them ; 
the  merchant,  not  being  able  to  make  the  change,  passes 
over  the  street  to  a  broker  and  gets  the  bill  changed,  and 
then  returns  and  gives  the  man,  who  bought  the  boots, 
his  change.  After  the  purchaser  of  the  boots  has  been 
gone  a  few  hours,  the  broker,  finding  the  bill  to  be  a 
counterfeit,  returns  and  demands  $50  of  good  money 
from  the  merchant.  How  much  did  the  merchant  lose 
by  the  operation? 

REMARK. — At  first  glance  some  say  $45  and  the  boots ;  some, 
$50  and  the  boots ;  some,  $95  and  the  boots ;  and  others,  $100 
and  the  boots.  Which  is  correct  ? 

14.  "What  relation  to  me  is  my  mother's  brother-in- 
law's  brother,  provided  he  has  but  one  brother  ? 

15.  Three  men,  travelling  with  their  wives,  came  to  a 
river  which  they  wished  to  cross.     There  was  but  one 
boat,  and  but  two  could  cross  at  one  time ;  and,  since  the 
husbands  were  jealous,  no  woman  could  be  with  a  man 
unless  her  own  husband  was  present.     In  what  manner 
did  they  get  across  the  river  ? 

REMARK. — Parke  states  that  this  problem  "  is  found  in  the 
works  of  Alcuin,  who  flourished  a  thousand  years  ago." 

16.  Suppose  it  were  possible  for  a  man,  in  Cincinnati, 
to  start  on  Sunday  noon,  when  the  sun  is  in  the  meridian, 
and  travel  westward  with  the  sun,  so  that  it  might  be  in 
his  meridian  all  the  time.    He  would  arrive  at  Cincinnati 
next  day  at  noon.     Now,  it  was  Sunday  noon  when  he 
started,  it  has  been  noon  with  him  all  the  way  around, 
and  is  Monday  noon  when  he  returns.     The  question  is, 


78  SOCIAL   ARITHMETIC. 

at  what  point  did  it  change  from  Sunday  noon  to  Monday 

noon  ? 

17.  Suppose  a  hare  is  10  rods  before  a  hound,  and 
that  the  hound  runs  10  rods  while  the  hare  runs  1  rod. 
Now  when  the  hound  has  run  the  10  rods,  the  hare  has 
run  1  rod ;  hence  they  are  now  1  rod  apart,  and  when 
the  hound  has  run  that  1  rod,  the  hare  has  run  y1^  of  a 
rod ;  hence  they  are  now  ^  of  a  rod  apart,  and  when 
the  hound  has  run  the  T^  of  a  rod,  they  are  y^  of  a  rod 
apart;  and  in  the  same  way  it  may  be  shown  the  hare  is 
always  y1^  of  the  previous  distance  ahead  of  the  hound; 
hence  the  hound  can  never  catch  the  hare.     How  is  the 
contrary  shown  mathematically  ? 

18.  Think  of  any  three  numbers  less  than  10.     Mul- 
tiply the  first  by  2,  and  add  5  to  the  product.     Multiply 
this  sum  by  5,  and  add  the  second  number  to  the  pro- 
duct.    Multiply  this  last  result  by  10,  and  add  the  third 
number  to  the  product ;  then  subtract  250.     Name  the 
remainder,  and  I  will  name  the  numbers  thought  of,  and 
in  the  order  in  which  they  were  thought  of. 

METHOD. — The  three  digits  composing  this  remainder,  will 
be  the  numbers  thought  of;  and  the  order  in  which  they  were 
thought  of  will  be  the  order  of  hundreds,  tens  and  units. 

19.  Write  24  with  three  equal  figures,  neither  of  them 
being  8. 

METHOD.— 22  -f  2  =  24,  or  3s  —  3  =  24. 

20.  Put  down  four  marks,  and  then  require  a  person 
to  put  down  five  more  marks,  and  make  ten. 

METHOD. — The  four  marks  are  as  represented         I     I     I  I 
in  the  margin;   the  five  more,  making  ten,  are 
placed  as  in  the  margin.  TEN 

21.  Which  is  the  greater,  and  how  much,  six  dozen 
dozen,  or  one-half  a  dozen  dozen,  or  is  there  no  difference 
between  them  ? 

22.  Show  what  is  wrong  in  the  following  reasoning : — 
8  —  8  equals  2  —  2 ;  dividing  both  these  equals  by  2 


SOCIAL   ARITHMETIC.  79 

-  2  and  the  results  must  be  equal;  8  —  8  divided  by 
£  -2  =  4,  and  2  —  2  divided  by  2  —  2  —  1;  there- 
fore, since  the  quotients  of  equals  divided  by  equals, 
must  be  equal,  4  must  be  equal  to  1. 

23.  A  man  has  a  triangular  lot  of  land,  the  largest 
Bide  being  136  rods,  and  each  of  the  other  sides  68  rods : 
required  the  value  of  the  grass  on  it,  at  the  rate  of  $10 
an  acre. 

REMARK. — The  "  catch"  in  this  is,  that  the  sides  given  will 
form  no  triangle. 

24.  Says  A  to  B,  "  Give  me  four  weights,  and  I  can 
weigh  any  number  of  pounds  not  exceeding  40."     Re- 
quired the  weights  and  the  method  of  weighing. 

ANSWER. — The  weights  are  1,  3,  9,  and  27  pounds.  In 
weighing,  we  must  put  one  or  more  in  both  scales,  or  some  iu 
one  scale  and  some  in  the  other;  thus,  7  Ibs.  =9  Ibs.  -j-1  Ib. 
—  3  Ibs. 

25.  Mr.  Frantz  planted  13  trees  in  his  garden,  in  such 
a  manner  that  there  were  12  rows,  and  only  3  trees  in 
each  row.     In  what  manner  were  they  planted  ? 

ANSWER. — They  were  in  the  form  of  a  regular  hexagon, 
having  a  tree  in  the  centre,  and  one  at  the  middle  and  extre- 
mity of  each  side. 

26.  A  and  B  raised  749  bushels  of  potatoes  on  shares ; 
A  was  to  have  |,  and  B  |  of  them.     Before  they  were 
divided,  however,  since  A  had  used  49  bushels,  B  took 
28  bushels  from  the  heap,  and  then  divided  the  remain- 
der according  to  the  above  agreement.     Was  this  divi- 
sion fair  ?  if  not,  show  how  it  should  have  been. 

27.  Two-thirds  of  six  is  nine,  one-half  of  twelve  ig 

seven, 

The  half  of  five  is  four,  and  six  is  half  of  eleven. 
SOLUTION. — Two-thirds  of  SIX  is  |X;    the  upper  half  of 
XII  is  VII;  the  half  of  FIVE  is  IV;  and  the  upper  half  of  X I 
is  VI- 

28.  Does  the  top  of  a  carriage-wheel  move  faster  than 
the  bottom  ?     If  so,  explain  it. 


80 


SOCIAL   ARITHMETIC. 


29.  Supposing  there  are  more  persons  in  the  world 
than  any  one  has  hairs  on  his  head,  there  must  be,  at 
least,  two  persons  who  have  the  same  number  of  hairs 
on  the  head,  to  a  hair.  Show  how  this  is. 

80.  Place  17  little  sticks — matches, 
for  instance — making  6  equal  squares, 
sis  in  the  margin.  Then  remove  5 
sticks,  and  leave  3  perfect  squares  of 
the  same  size. 

31.  Three  persons  own  51  quarts  of  rice,  and  have 
only  two  measures ;  one  a  4  quart,  the  other  a  7  quart 
measure.  How  shall  they  divide  it  into  three  equal 

parts  ? 

METHOD. — Perhaps  the  easiest  way  is  to  give  each  one  17 
quarts,  which  may  be  obtained  thus :  fill  the  7  quart  measure ; 
empty  this  into  the  4  quart  measure,  and  there  will  be  3  quarts 
in  the  7  quart  measure,  which  added  to  two  7  quart  measures, 
equals  17  quarts. 

82.  What  four  United  States  coins  will  amount  to 
fifty-one  cents  ? 

ANSWER. — Two  25  ct.  pieces  and  two  half-cents. 

33.  How  may  the  nine  digits  be 
arranged  in  a  rectangular  form,  so 
that  the  sum  of  any  row,  whether 
horizontal,  vertical,  or  diagonal,  shall 
equal  15  ? 

ANSWER. — As  in  the  margin. 


34.  How  may  the  first  16  di- 
gits be  arranged,  so  that  the  sum 
of  the  vertical,  the  horizontal, 
and  the  two  oblique  rows  may 
«qual  thirty-four  ?  % 

ANSWER. — As  .in  the  margin. 


1 

16 

11 

6  i 

13 

4 

7 

10  | 

8 

9 

14 

3    i 

12 

5 

2 

15 

SOCIAL   ARITHMETIC. 


81 


1 

10 

12 

18 

24 

9 

11 

20 

22 

3 

13 

19 

21 

5 

7 

17 

23 

4 

6 
14 

15 

25 

2 

8 

16 

35.  In  what  manner  may 
the   first   25    digits   be   ar- 
ranged, so  that  the  sum  of 
each  row  of  five  figures  may 
be  65  ? 

ANSWER. — As  in  the  margin. 

REMARK. — The  above  are 
i-alled  Magic  Squares.  They 
are  very  interesting,  and  have 
engaged  the  attention  of  some 
of  our  greatest  mathematicians, 
among  whom  we  may  mention  Leibnitz,  Stifels,  &c.  The 
methods  of  arrangement  given  above  are  by  no  means  the  only 
ones  that  may  be  used.  For  the  second  problem,  Frenicle,  a 
French  mathematician,  has  shown  that  there  may  be  878  differ- 
ent arrangements. 

36.  Take  10  pieces  of  money,  lay  them  in  a  row,  and 
require  some  one  to  put  them  together  in  heaps  2  in  each, 
by  passing  each  piece  over  %  others. 

METHOD. — Let  the  pieces  be  represented  by  the  numbers  1, 
2,  3,  4,  5,  6,  7,  8,  9,  10.  Place  7  on  10,  5  on  2,  3  on  8,  1  on  4, 
and  9  on  6. 

37.  An  old  Jew  took  a  diamond  cross  to  a  jeweller,  to 
have  the  diamonds  reset;  and  fearing  that  the  jeweller 
might  be  dishonest,  he  counted  the  diamonds,  and  found 
that  they  numbered  7  in  three  different  ways.     Now  the 
jeweller  stole  two  diamonds,  but  arranged  the  remainder 
so  that  they  counted  7  each  way,  as  before.     How  was 
it  done  ? 

METHOD. — The  form  of  the  cross  when  left  **•  L  Fig.  2. 
is  represented  by  Fig.  1,  and  when  returned  *  y 

by  Fig.  2.     It  will  be  seen  by  the  figures  how  ^6567 
the  diamonds  were  counted  by  the  old  Jew,  and       4:  4= 

how  they  were  arranged  by  the  jeweller,  who        3 
"jewed"  the  Jew.  j 

38.  Let  a  person  select  a  number  greater  than  1  and 
not  exceeding  10 ;  I  will  add  to  it  a  number  not  exceed- 
ing 10,  alternately  with  himself;  and,  although  he  ha* 


82  SOCIAL   ARITHMETIC. 

the  advantage  in  selecting  the  number  to  start  with,  I 
will  reach  the  even  hundred  first. 

METHOD. — I  make  my  additions  so  that  the  sums  are,  re- 
spectively, 12,  23,  34,  45,  &c.,  to  89,  when  it  is  evident  I  can 
reach  the  hundred  first.  With  one  who  does  not  mistrust  the 
method,  I  need  not  run  through  the  entire  series,  but  merely 
aim  for  89,  or,  when  the  secret  of  this  is  seen,  for  78,  then 
t57,  &c. 

39.  Let  a  person  think  of  any  number  on  the  dial-face 
of  a  watch ;  I  will  then  point  to  various  numbers,  and 
at  each  he  will  silently  add  one  to  the  number  selected, 
until  he  arrives  at  twenty,  which  he  will  announce  aloud, 
and  my  pointer  will  be  upon  the  number  he  selected. 

METHOD. — I  point  promiscuously  about  the  face  of  the  watch 
until  the  eighth  point,  which  should  be  upon  "12  ;"  and  then 
pass  regularly  around  towards  "1,"  pointing  at  "11,"  "10," 
"9,"  &c.,  until  "twenty"  is  called,  when,  as  may  be  easily 
shown,  my  pointer  will  be  over  the  number  selected. 

40.  Is  there  any  difference  between  the  results  of  the 
two  following  problems,  and*  if  so,  what  is  it  ?     If  the 
half  of  6  be  3,  what  will  the  fourth  of  20  be  ?     If  3  be 
the  half  of  6,  what  will  be  the  fourth  of  20  ? 

41.  A  vessel  with  a  crew  of  30  men,  half  of  whom 
were  black,  became  short  of  provisions;    and,  fearing 
that  unless  half  the  crew  were  thrown  overboard  all 
would  perish,  the  captain  proposed  to  the  sailors  to  stand 
upon  deck  in  a  row,  and  every  ninth  man  be  thrown 
overboard  until  half  the  crew  were   destroyed.     It  so 
happened  that  the  whites  were  saved.      Required  the 
order  of  arrangement. 

ANSWER.— W.  W.  W.  W.  B.  B.  B.  B.  B.  W.  W.  B.  W.  W.  W. 
B.  W.  B.  B.  W.  W.  B.  B.  B.  W.  B.  B.  W.  W.  B. 

This  can  easily  be  found  by  trial,  using  letters  or  figures  to 
represent  the  men. 

42.  Think  of  a  number,  multiply  it  by  6,  divide  this 
product  by  2,  multiply  by  4,  divide  by  3,  add  40,  divide 
by  4,  subtract  the  number  thought  of,  divide  by  2,  and 
the  quotient  is  5.     Show  why  this  is  so. 


SOCIAL    ARITHMETIC.  S3 

'  43.  A  and  B  were  engaged  by  a  Chester  Co.  farmer 
to  dig  100  rods  of  ditch  for  $100;  and  since  the  part 
which  A  was  to  dig  was  more  difficult  of  excavation  than 
that  which  B  dug,  it  was  agreed  that  A  should  receive 
10  shillings  per  rod,  and  B  6  shillings  per  rod.  They 
each  received  $50  for  their  labor.  How  many  rods  did 
each  dig  ? 

REMARK. — This  problem  is  usually  regarded  as  incapable  of 
solution,  incorrectly  so,  however ;  for  the  results  obtained  by 
a  mathematical  process,  rigidly  accurate,  are,  A  dug  37J,  and 
B  62J  rods. 

The  majority  of  these  puzzles  and  problems  being 
founded  upon  principles  quite  easily  comprehended,  the 
author  has  not  thought  it  necessary  to  explain  the  prin- 
ciples of  the  puzzles  nor  solve  the  problems.  It  is  hoped 
that  they  may  prove  a  source  of  pleasure  and  profit  to 
teacher  and  pupil 


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New  edition^  and  exceedin 


PULTON'S  OUTLINE  MAPS. 
This  Series  of  six  superb  M;    •    is  pnw  adopted  in  n)mo.«i  ;wrry  fchoo}  of  note 
in  the  Union  where  pea?  hf.  ;ind   HAS  NO  WALS,      4 

PHYSICAL  QEOGRApnr  has  been  introduced  upon  the  H»>rii?)>here  M*ps, 
its  detail."  having  been  so  managed  15  to  present  a  most  ^-r^eous  nppraran 
while  ther  do  not  interfere  in   ihe  lea.«t  with,  e^ch  otiier,  pif  the   subjects 
usunUy  shown  \\pon  Maps. 

A  new  Map,  of  the  United  States  hns  ntso  been  cub^tituted,  including  the 
whole  of  North  America,  from,  Labrador  to  the  lsth,u»vw. 

A  complete  ,*tork  r.f  School  Book  P.  published  in  this  and  other  cities,  con- 
stantly 011  huu,4,  t0gather  with  Miscellaneous  Uo.»k?  and  Stationery. 

SOWER,  BARNES  &  CO., 
37  North  Third  Street,  Philadelphia. 


